Miscellaneous Exercise : Proofs of Trigonometric Identities

1. Prove that $\dfrac{\tan \theta+\cot \theta}{\sec \theta+\operatorname{cosec} \theta}=\dfrac{1}{\sin \theta+\cos \theta}$





$\begin{aligned} &\dfrac{\tan \theta+\cot \theta}{\sec \theta+\csc \theta}\\\\ = &\dfrac{\dfrac{\sin \theta}{\cos \theta}+\dfrac{\cos \theta}{\sin \theta}}{\dfrac{1}{\cos \theta}+\dfrac{1}{\sin \theta}}\\\\ = &\dfrac{\dfrac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin \theta \cos \theta}}{\dfrac{\sin \theta+\cos \theta}{\sin \theta \cos \theta}}\\\\ = &\dfrac{1}{\sin \theta \cos \theta} \times \dfrac{\sin \theta \cos \theta}{\sin \theta+\cos \theta}\\\\ = &\dfrac{1}{\sin \theta+ \cos \theta} \end{aligned}$



2. Prove that $\quad \sec x \csc x-\cot x=\tan x$.





$\begin{aligned} & \sec x \csc x-\cot x \\\\ = & \dfrac{1}{\cos x} \dfrac{1}{\sin x}-\dfrac{\cos x}{\sin x} \\\\ = & \dfrac{1-\cos ^{2} x}{\sin x \cos x} \\\\ = & \dfrac{\sin ^{2} x}{\sin x \cos x} \\\\ = & \dfrac{\sin x}{\cos x} \\\\ = & \tan x \end{aligned}$



3. Prove that $\dfrac{1}{1-\cos x}+\dfrac{1}{1+\cos x}=2 \csc^{2} x$





$\begin{aligned} & \dfrac{1}{1-\cos x}+\dfrac{1}{1+\cos x} \\\\ = & \dfrac{1+\cos x+1-\cos x}{(1-\cos x)(1+\cos x)} \\\\ = & \dfrac{2}{1-\cos ^{2} x} \\\\ = & \dfrac{2}{\sin ^{2} x} \\\\ = & 2 \csc ^{2} x \end{aligned}$



4. Show that $\dfrac{\tan \theta+\cot \theta}{\csc \theta}=\sec \theta$





$\begin{aligned} &\dfrac{\tan \theta+\cot \theta}{\csc \theta}\\\\ = &\dfrac{\dfrac{\sin \theta}{\cos \theta}+\dfrac{\cos \theta}{\sin \theta}}{\dfrac{1}{\sin \theta}}\\\\ = &\dfrac{\dfrac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin \theta \cos \theta}}{\dfrac{1}{\sin \theta}}\\\\ = &\dfrac{1}{\sin \theta \cos \theta} \times \sin \theta \\\\ = &\sec \theta \end{aligned}$



5. Show that $\sqrt{\sec ^{2} \theta-1}+\sqrt{\csc^{2} \theta-1}=\sec \theta \csc \theta$





$\begin{aligned} & \sqrt{\sec ^{2} \theta-1}+\sqrt{\csc ^{2} \theta-1} \\\\ =& \sqrt{\tan ^{2} \theta}+\sqrt{\cot ^{2} \theta} \\\\ =& \tan \theta+\cot \theta \\\\ =& \dfrac{\sin \theta}{\cos \theta}+\dfrac{\cos \theta}{\sin \theta} \\\\ =& \dfrac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin \theta \cos \theta} \\\\ =& \dfrac{1}{\sin \theta \cos \theta} \\\\ =& \sec \theta \csc \theta \end{aligned}$



6. Prove that $\sec ^{2} x+\csc^{2} x=\sec ^{2} x \csc^{2} x$





$\begin{aligned} & \sec ^{2} x+\csc ^{2} x \\\\ =& \dfrac{1}{\cos ^{2} x}+\dfrac{1}{\sin ^{2} x} \\\\ =& \dfrac{\sin ^{2} x+\cos ^{2} x}{\sin ^{2} x \cos ^{2} x} \\\\ =& \dfrac{1}{\sin ^{2} x \cos ^{2} x} \\\\ =& \dfrac{1}{\cos ^{2} x} \cdot \dfrac{1}{\sin ^{2} x}\\\\ =&\sec ^{2} x \csc ^{2} x \end{aligned}$



7. Show that $\dfrac{1}{\csc \theta-1}-\dfrac{1}{\csc \theta+1}=2 \tan ^{2} \theta$





$\begin{aligned} & \dfrac{1}{\csc \theta-1}-\dfrac{1}{\csc \theta+1} \\\\ =& \dfrac{(\csc \theta+1)-(\csc \theta-1)}{(\csc \theta-1)(\csc \theta+1)} \\\\ =& \dfrac{2}{\csc ^{2} \theta-1} \\\\ =& \dfrac{2}{\cot ^{2} \theta} \\\\ =& 2 \tan ^{2} \theta \end{aligned}$



8. Show that $\dfrac{\csc \theta}{\csc \theta-\sin \theta}=\sec ^{2} \theta$.





$\begin{aligned} &\dfrac{\csc \theta}{\csc \theta-\sin \theta} \\\\ =& \dfrac{\dfrac{1}{\sin \theta}}{\dfrac{1}{\sin \theta}-\sin \theta} \\\\ =& \dfrac{1}{\sin \theta} \times \dfrac{\sin \theta}{\cos ^{2} \theta}\\\\ =&\sec ^{2} \theta \end{aligned}$



9. Prove that $\dfrac{\cos x}{1+\tan x}-\dfrac{\sin x}{1+\cot x}=\cos x-\sin x$.





$\begin{aligned} &\dfrac{\cos x}{1+\tan x}-\dfrac{\sin x}{1+\cot x} \\\\ =&\dfrac{\cos x}{1+\dfrac{\sin x}{\cos x}}-\dfrac{\sin x}{1+\dfrac{\cos x}{\sin x}} \\\\ =&\dfrac{\cos x+\sin x}{\cos x}-\dfrac{\sin x}{\dfrac{\sin x+\cos u}{\sin x}}\\\\ =&\cos x\left(\dfrac{\cos x}{\cos x+\sin x}\right)-\sin x\left(\dfrac{\sin x}{\sin x+\cos x}\right)\\\\ =&\dfrac{\cos ^{2} x-\sin ^{2} x}{\cos x+\sin x}\\\\ =&\dfrac{(\cos x-\sin x)(\cos x+\sin x)}{\cos x+\sin x}\\\\ =&\cos x-\sin x \end{aligned}$



10. Show that $\csc \theta-\sin \theta=\cot \theta \cos \theta$.





$\begin{aligned} & \csc \theta-\sin \theta \\\\ =& \dfrac{1}{\sin \theta}-\sin \theta \\\\ =& \dfrac{1-\sin ^{2} \theta}{\sin \theta} \\\\ =& \dfrac{\cos ^{2} \theta}{\sin \theta} \\\\ =& \dfrac{\cos \theta}{\sin \theta} \cos \theta \\\\ =& \cot \theta \cos \theta \end{aligned}$



11. Show that $\dfrac{\tan ^{2} \theta+\sin ^{2} \theta}{\cos \theta+\sec \theta}=\tan \theta \sin \theta$.





$\begin{aligned} & \dfrac{\tan ^{2} \theta+\sin ^{2} \theta}{\cos \theta+\sec \theta} \\\\ =& \dfrac{\dfrac{\sin ^{2} \theta}{\cos ^{2} \theta}+\sin ^{2} \theta}{\cos \theta+\dfrac{1}{\cos \theta}} \\\\ =& \dfrac{\sin ^{2} \theta\left(1+\cos ^{2} \theta\right)}{\cos ^{2} \theta} \times \dfrac{\cos \theta}{\cos ^{2} \theta+T}\\\\ =&\dfrac{\sin ^{2} \theta}{\cos \theta} \\\\ =&\dfrac{\sin \theta}{\cos \theta} \sin \theta \\\\ =&\tan \theta \sin \theta \end{aligned}$



12. Prove that $\sin x(\cot x+\tan x)=\sec x$





$\begin{aligned} & \sin x(\cot x+\tan x) \\ =& \sin x\left(\dfrac{\cos x}{\sin x}+\dfrac{\sin x}{\cos x}\right) \\\\ =& \sin x\left(\dfrac{\cos 2 x+\sin ^{2} x}{\sin x \cos x}\right) \\\\ =& \dfrac{1}{\cos x} \\\\ =& \sec x \end{aligned}$



13. Show that $\dfrac{\sec \theta}{\cot \theta+\tan \theta}=\sin \theta$.





$\begin{aligned} &\dfrac{\sec \theta}{\cot \theta+\tan \theta}\\\\ =&\dfrac{\dfrac{1}{\cos \theta}}{\dfrac{\cos \theta}{\sin \theta}+\dfrac{\sin \theta}{\cos \theta}}\\\\ =&\dfrac{\dfrac{1}{\cos \theta}}{\dfrac{\cos ^{2} \theta+\sin ^{2} \theta}{\sin \theta \cos \theta}}\\\\ =&\dfrac{1}{\cos \theta} \times \dfrac{\sin \theta \cos \theta}{1}\\\\ =&\sin \theta \end{aligned}$



14. Show that $\dfrac{\sin x}{1+\cos x}+\dfrac{1+\cos x}{\sin x}=2 \csc x$.





$\begin{aligned} & \dfrac{\sin x}{1+\cos x}+\dfrac{1+\cos x}{\sin x} \\\\ =& \dfrac{\sin x}{1+\cos x} \times \dfrac{1-\cos x}{1-\cos x}+\dfrac{1+\cos x}{\sin x} \\\\ =& \dfrac{\sin x(1-\cos x)}{1-\cos ^{2} x}+\dfrac{1+\cos x}{\sin x} \\\\ =& \dfrac{\sin x(1-\cos x)}{\sin ^{2} x}+\dfrac{1+\cos x}{\sin x}\\\\ =&\dfrac{1-\cos x}{\sin x}+\dfrac{1+\cos x}{\sin x} \\\\ =&\dfrac{1-\cos x+1+\cos x}{\sin x} \\\\ =&\dfrac{2}{\sin x} \\\\ =&2 \csc x \end{aligned}$



15. Show that $\dfrac{(1-\sin A)(1+\sin A)}{\sin A \cos A}=\cot A$.





$\begin{aligned} & \dfrac{(1-\sin A)(1+\sin A)}{\sin A \cos A} \\\\ =& \dfrac{1-\sin ^{2} A}{\sin A \cos A} \\\\ =& \dfrac{\cos ^{2} A}{\sin A \cos A} \\\\ =& \dfrac{\cos A}{\sin A}\\\\ =&\cot A \end{aligned}$



16. Show that $\cos \theta \cot \theta+\sin \theta=\csc \theta$.





$\begin{aligned} & \cos \theta \cot \theta+\sin \theta \\\\ =& \cos \theta \cdot \dfrac{\cos \theta}{\sin \theta}+\sin \theta \\\\ =& \dfrac{\cos ^{2} \theta}{\sin \theta}+\sin \theta \\\\ =& \dfrac{\cos ^{2} \theta+\sin ^{2} \theta}{\sin \theta} \\\\ =& \dfrac{1}{\sin \theta} \\\\ =& \csc \theta \end{aligned}$



17. Show that $2 \cos x \cot x+1=\cot x+2 \cos x$ can be written in the form $(a \cos x-b)(\cos x-\sin x)=0$, where $a$ and $b$ are constants to be found.





$\begin{aligned} &\quad 2 \cos x \cot x+1=\cot x+2 \cos x \\\\ &\quad 2 \cos x \dfrac{\cos x}{\sin x}+1=\dfrac{\cos x}{\sin x}+2 \cos x \\\\ &\quad \text { Multiplying both sides with}\ \sin x, \\\\ &\quad 2 \cos ^{2} x+\sin x=\cos x+2 \sin x \cos x \\\\ &\quad 2 \cos ^{2} x-\cos x+\sin x-2 \sin x \cos x=0 \\\\ &\quad\cos x(2 \cos x-1)+\sin x(1-2 \cos x)=0\\\\ &\quad\cos x(2 \cos x-1)-\sin x(2 \cos x-1)=0 \\\\ &\quad(2 \cos x-1)(\cos x-\sin x)=0 \\\\ &\therefore(a \cos x-b)(\cos x-\sin x)=(2 \cos x-1)(\cos x-\sin x) \\\\ &\therefore a=2\quad \text { and }\quad b=1 \end{aligned}$

Proving Trigonometric Identities : Exercise (10.3) - Solutions

Basic Trigonometric Identities

$\begin{array}{|l|} \hline \sin A=\displaystyle \frac{1}{\csc A}\ \text{and}\\ \csc A=\displaystyle \frac{1}{\sin A}\\ \cos A=\displaystyle \frac{1}{\sec A}\ \text{and}\\ \sec A=\displaystyle \frac{1}{\cos A}\\ \tan A=\displaystyle \frac{\sin A}{\cos A}\\ \tan A=\displaystyle \frac{1}{\cot A}\ \text{and}\\ \cot A=\displaystyle \frac{1}{\tan A}\\ \cot A=\displaystyle \frac{\cos A}{\sin A}\\ \hline\end{array}$

Pythagorean Identities

$\begin{array}{|l|} \hline \sin^2 A+ \cos^2 A=1\\ \tan^2 A+ 1=\sec^2 A\\ 1+ \cot^2 A=\csc^2 A\\ \hline\end{array}$

Trigonometric Ratios of Complementary Angles

$\begin{array}{|l|}\hline \sin \left(90^{\circ}-\alpha\right)=\cos \alpha\\ \cos \left(90^{\circ}-\alpha\right)=\sin \alpha\\ \tan \left(90^{\circ}-\alpha\right)=\cot \alpha\\ \cot \left(90^{\circ}-\alpha\right)=\tan \alpha\\ \sec \left(90^{\circ}-\alpha\right)=\csc \alpha\\ \csc \left(90^{\circ}-\alpha\right)=\sec \alpha\\ \hline \end{array}$



Prove the following identities.

1.           $\cot \theta \sqrt{1-\cos ^{2} \theta}=\cos \theta$

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$$\begin{aligned} &\ \ \ \ \ \cot \theta \sqrt{1-\cos ^{2} \theta}\\\\ &=\cot \theta \sqrt{\sin ^{2} \theta}\\\\ &=\frac{\cos \theta}{\sin \theta} \times \sin \theta\\\\ &=\cos \theta\end{aligned}$$

2.           $\displaystyle \frac{\tan ^{2} \theta+1}{\tan \theta \csc ^{2} \theta}=\tan \theta$

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$$\begin{aligned} &\ \ \ \ \ \frac{\tan ^{2} \theta+1}{\tan \theta \csc ^{2} \theta}\\\\ &=\frac{\sec ^{2} \theta}{\cos \theta} \times \frac{1}{\sin ^{2} \theta}\\\\ &=\frac{1}{\cos ^{2} \theta} \times \sin \theta \cos \theta\\\\ &=\frac{\sin \theta}{\cos \theta}=\tan \theta \end{aligned}$$

3.           $\left(1-\sin ^{2} \theta\right)\left(1+\cot ^{2} \theta\right)=\cot ^{2} \theta$

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$\begin{aligned} &\ \ \ \ \ \left(1-\sin ^{2} \theta\right)\left(1+\cot ^{2} \theta\right)\\\\ &=\cos ^{2} \theta \csc ^{2} \theta\\\\ &=\frac{\cos ^{2} \theta}{\sin ^{2} \theta}\\\\ &=\cot ^{2} \theta\end{aligned}$

4.           $\tan ^{2} \theta-\cot ^{2} \theta=\sec ^{2} \theta-\csc ^{2} \theta$

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$\begin{aligned} &\ \ \ \ \ \tan ^{2} \theta-\cot ^{2} \theta\\\\ & =\left(\sec ^{2} \theta-1\right)-\left(\csc ^{2} \theta-1\right)\\\\ &=\sec ^{2} \theta-1-\csc ^{2} \theta+1\\\\ &=\sec ^{2} \theta-\csc ^{2} \theta\end{aligned}$

5.           $\sin \theta \sec \theta \sqrt{\csc ^{2} \theta-1}=1$

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$\begin{aligned} &\ \ \ \ \ \sin \theta \sec \theta \sqrt{\csc ^{2} \theta-1}\\\\ &=\frac{\sin \theta}{\cos \theta} \times \sqrt{\cot ^{2} \theta}\\\\ &=\tan \theta \cot \theta=1\end{aligned}$

6.           $16 \sec ^{2} \theta+\csc ^{2} \theta=\sec ^{2} \theta \csc ^{2} \theta$

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$\begin{aligned} &\ \ \ \ \ \sec ^{2} \theta+\csc ^{2} \theta \\\\ &=\frac{1}{\cos ^{2} \theta}+\frac{1}{\sin ^{2} \theta}\\\\ &=\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\cos ^{2} \theta \sin ^{2} \theta}\\\\ &=\frac{1}{\cos ^{2} \theta \sin ^{2} \theta}\\\\ &=\sec ^{2} \theta \csc ^{2} \theta\end{aligned}$

7.           $\left(1+\tan ^{2} \theta\right)\left(1-\sin ^{2} \theta\right)=1$

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$\begin{aligned} &\ \ \ \ \ \left(1+\tan ^{2} \theta\right)\left(1-\sin ^{2} \theta\right)\\\\ &=\sec ^{2} \theta \cos ^{2} \theta\\\\ &=\frac{1}{\cos ^{2} \theta} \times \cos ^{2} \theta\\\\ &=1\end{aligned}$

8.           $(1+\tan \theta)^{2}+(1-\tan \theta)^{2}=2 \sec ^{2} \theta$

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$\begin{aligned} &\ \ \ \ \ (1+\tan \theta)^{2}+(1-\tan \theta)^{2}\\\\ &=\left(1+2 \tan \theta+\tan ^{2} \theta\right)+\left(1-2 \tan \theta+\tan ^{2} \theta\right)\\\\ &=2+2 \tan ^{2} \theta\\\\ &=2\left(1+\tan ^{2} \theta\right)\\\\ &=2 \sec ^{2} \theta\end{aligned}$

9.           $\sec ^{2} \theta \cot ^{2} \theta-1=\cot ^{2} \theta$

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$\begin{aligned} &\ \ \ \ \ \sec ^{2} \theta \cot ^{2} \theta-1 \\\\ &=\frac{1}{\cos ^{2} \theta} \times \frac{\cos ^{2} \theta}{\sin ^{2} \theta}-1\\\\ &=\frac{1}{\sin ^{2} \theta}-1\\\\ &=\csc ^{2} \theta-1\\\\ &=\cot ^{2} \theta\end{aligned}$

10.           $\displaystyle \frac{1}{1-\sin \theta}+\displaystyle \frac{1}{1+\sin \theta}=2 \sec ^{2} \theta$

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$\begin{aligned} &\ \ \ \ \ \frac{1}{1-\sin \theta}+\frac{1}{1+\sin \theta}\\\\ &=\frac{1+\sin \theta+1-\sin \theta}{(1-\sin \theta)(1+\sin \theta)}\\\\ &=\frac{2}{1-\sin ^{2} \theta}\\\\ &=\frac{2}{\cos ^{2} \theta}\\\\ &=2 \sec ^{2} \theta\end{aligned}$

11.           $\displaystyle \frac{1}{\sin ^{2} \theta}-\displaystyle \frac{1}{\tan ^{2} \theta}=1$

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$\begin{aligned} &\ \ \ \ \ \frac{1}{\sin ^{2} \theta}-\frac{1}{\tan ^{2} \theta}\\\\ &=\csc ^{2} \theta-\cot ^{2} \theta=1\end{aligned}$

12.           $(\tan \theta+\sec \theta)^{2}=\displaystyle \frac{1+\sin \theta}{1-\sin \theta}$

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$\begin{aligned} &\ \ \ \ \ (\tan \theta+\sec \theta)^{2}\\\\ &=\left(\frac{\sin \theta}{\cos \theta}+\frac{1}{\cos \theta}\right)^{2}\\\\ &=\left(\frac{1+\sin \theta}{\cos \theta}\right)^{2}\\\\ &=\frac{(1+\sin \theta)^{2}}{\cos ^{2} \theta}\\\\ &=\frac{(1+\sin \theta)^{2}}{1-\sin ^{2} \theta}\\\\ &=\frac{(1+\sin \theta)^{2}}{(1-\sin \theta)(1+\sin \theta)}\\\\ &=\frac{1+\sin \theta}{1-\sin \theta}\end{aligned}$

13.           $\sin ^{4} \theta-\cos ^{4} \theta=1-2 \cos ^{2} \theta$

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$\begin{aligned} &\ \ \ \ \ \sin ^{4} \theta-\cos ^{4} \theta \\\\ &=\left(\sin ^{2} \theta\right)^{2}-\left(\cos ^{2} \theta\right)^{2}\\\\ &=\left(\sin ^{2} \theta+\cos ^{2} \theta\right)\left(\sin ^{2} \theta-\cos ^{2} \theta\right) \\\\ &=1\left(1-\cos ^{2} \theta-\cos ^{2} \theta\right)\\\\ &=1-2 \cos ^{2} \theta\end{aligned}$

14.           $\displaystyle \frac{\tan ^{2} \theta+1}{\tan ^{2} \theta}=\csc ^{2} \theta$

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$\begin{aligned} &\ \ \ \ \ \frac{\tan ^{2} \theta+1}{\tan ^{2} \theta}\\\\ &=\frac{\tan ^{2} \theta}{\tan ^{2} \theta}+\frac{1}{\tan ^{2} \theta}\\\\ &=1+\cot ^{2} \theta\\\\ &=\csc ^{2} \theta\end{aligned}$

15.           $\sin ^{2} \theta \tan \theta+\cos ^{2} \theta \cot \theta+2 \sin \theta \cos \theta=\tan \theta+\cot \theta$

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$\begin{aligned} &\ \ \ \ \ {{\sin }^{2}}\theta \tan \theta +{{\cos }^{2}}\theta \cot \theta +2\sin \theta \cos \theta \\ &={{\sin }^{2}}\theta \cdot \frac{{\sin \theta }}{{\cos \theta }}+{{\cos }^{2}}\theta \cdot \frac{{\cos \theta }}{{\sin \theta }}+2\sin \theta \cos \theta \\ &=\frac{{{{{\sin }}^{3}}\theta }}{{\cos \theta }}+\frac{{{{{\cos }}^{3}}\theta }}{{\sin \theta }}+2\sin \theta \cos \theta \\ &=\frac{{{{{\sin }}^{4}}\theta +{{{\cos }}^{4}}\theta +2{{{\sin }}^{2}}\theta {{{\cos }}^{2}}\theta }}{{\sin \theta \cos \theta }}\\ &=\frac{{{{{\left( {{{{\sin }}^{2}}\theta +{{{\cos }}^{2}}\theta } \right)}}^{2}}}}{{\sin \theta \cos \theta }}\\ &=\frac{{\left( {{{{\sin }}^{2}}\theta +{{{\cos }}^{2}}\theta } \right)\left( {{{{\sin }}^{2}}\theta +{{{\cos }}^{2}}\theta } \right)}}{{\sin \theta \cos \theta }}\\ &=\frac{{{{{\sin }}^{2}}\theta +{{{\cos }}^{2}}\theta }}{{\sin \theta \cos \theta }}\\ &=\frac{{{{{\sin }}^{2}}\theta }}{{\sin \theta \cos \theta }}+\frac{{{{{\cos }}^{2}}\theta }}{{\sin \theta \cos \theta }}\\ &=\tan \theta +\cot \theta \end{aligned}$

16.           Find the value of acute angle $\alpha$ in each of the following equations:

               (a)    $\cos 2 \alpha=\sin 7 \alpha$

               (b)    $\tan 3 \alpha=\cot 2 \alpha$

               (c)    $\sec \alpha=\csc 5 \alpha$

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$\begin{array}{l} \text {(a)} \quad \cos 2 \alpha=\sin 7 \alpha\\\\ \therefore \quad 2\alpha + 7 \alpha = 90^{\circ}\\\\ \quad\quad 9\alpha= 90^{\circ}\\\\ \therefore \quad \alpha= 90^{\circ}\\\\ \text {(b)} \quad \tan 3 \alpha=\cot 2 \alpha\\\\ \therefore \quad 3\alpha + 2 \alpha = 90^{\circ}\\\\ \quad\quad 5\alpha= 90^{\circ}\\\\ \therefore \quad \alpha= 18^{\circ}\\\\ \text {(c)} \quad \sec \alpha=\csc 5 \alpha\\\\ \therefore \quad \alpha + 5 \alpha = 90^{\circ}\\\\ \quad\quad 6\alpha= 90^{\circ}\\\\ \therefore \quad \alpha= 15^{\circ} \end{array}$

17.           Prove the identity $\cos \left(90^{\circ}-\alpha\right) \tan \left(90^{\circ}-\alpha\right)=\cos \alpha$

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$\begin{aligned} &\ \ \ \ \ \cos (90-\alpha) \tan (90-\alpha)\\\\ &=\sin \alpha \cot \alpha\\\\ &=\sin \alpha \times \frac{\cos \alpha}{\sin \alpha}\\\\ &=\cos \alpha \end{aligned}$

18.           Prove the identity $\sin \left(90^{\circ}-\alpha\right) \sec \left(90^{\circ}-\alpha\right)=\cot \alpha$

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$\begin{aligned} &\ \ \ \ \ \sin (90-\alpha) \sec \left(90^{\circ}-\alpha\right)\\\\ &=\cos \alpha \csc \alpha\\\\ &=\cos \alpha \times \frac{1}{\sin \alpha}\\\\ &=\frac{\cos \alpha}{\sin \alpha}\\\\ &=\cot \alpha\end{aligned}$

Trigonometry : Practice Problems

1.        If $ \displaystyle \tan \alpha + \tan \beta = a$, $ \displaystyle \cot \alpha + \cot \beta = b$ and $ \displaystyle \tan(\alpha +\beta) = c$, show that $ \displaystyle bc - ac = ab$.

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$ \displaystyle \begin{array}{l}\ \ \ \ \tan \alpha +\tan \beta =a\\\\\ \ \ \ \cot \alpha +\cot \beta =b\\\\\therefore \ \ \displaystyle \frac{1}{{\tan \alpha }}+\displaystyle \frac{1}{{\tan \beta }}=b\\\\\therefore \ \ \displaystyle \frac{{\tan \alpha +\tan \beta }}{{\tan \alpha \tan \beta }}=b\\\\\therefore \ \ \displaystyle \frac{a}{{\tan \alpha \tan \beta }}=b\\\\\therefore \ \ \tan \alpha \tan \beta =\displaystyle \frac{a}{b}\\\\\ \ \ \ \tan \left( {\alpha +\beta } \right)=c\\\\\ \ \ \ \displaystyle \frac{{\tan \alpha +\tan \beta }}{{1-\tan \alpha \tan \beta }}=c\\\\\therefore \ \ \displaystyle \frac{a}{{1-\displaystyle \frac{a}{b}}}=c\\\\\therefore \ \ \displaystyle \frac{{ab}}{{b-a}}=c\\\\\therefore \ \ ab=bc-ac\end{array}$

2.        If $\displaystyle \alpha +\beta =45{}^\circ $, find the value of $ \displaystyle (1+\tan \alpha )(1+\tan \beta )$.

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \alpha +\beta =45{}^\circ \\\\\therefore \ \ \ \ \tan \left( {\alpha +\beta } \right)=\tan 45{}^\circ \\\\\ \ \ \ \ \displaystyle \frac{{\tan \alpha +\tan \beta }}{{1-\tan \alpha \tan \beta }}=1\\\\\ \ \ \ \ \tan \alpha +\tan \beta =1-\tan \alpha \tan \beta \\\\\therefore \ \ \ \tan \alpha +\tan \beta +\tan \alpha \tan \beta =1\\\\\therefore \ \ \ 1+\tan \alpha +\tan \beta +\tan \alpha \tan \beta =2\\\\\therefore \ \ \ \left( {1+\tan \alpha } \right)+\tan \beta \left( {1+\tan \alpha } \right)=2\\\\\therefore \ \ \ \left( {1+\tan \alpha } \right)\left( {1+\tan \beta } \right)=2\end{array}$

3.        If $ \displaystyle \cos \alpha +\cos \beta +\cos \gamma =0$ and $ \displaystyle \sin \alpha +\sin \beta +\sin \gamma =0$, find the value of $ \displaystyle \cos \left( {\alpha -\beta } \right)+\cos \left( {\beta -\gamma } \right)+\cos \left( {\gamma - \alpha} \right)$.

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$ \displaystyle \begin{array}{l}\ \ \ \ \cos \alpha +\cos \beta +\cos \gamma =0\\\\\ \ \ \ \sin \alpha +\sin \beta +\sin \gamma =0\\\\\therefore \ \ {{\left( {\sin \alpha +\sin \beta +\sin \gamma } \right)}^{2}}+{{\left( {\cos \alpha +\cos \beta +\cos \gamma } \right)}^{2}}=0\\\\\ \ \ \ {{\sin }^{2}}\alpha +{{\sin }^{2}}\beta +{{\sin }^{2}}\gamma +2\left( {\sin \alpha \sin \beta +\sin \beta \sin \gamma +\sin \alpha \sin \gamma } \right)\ \ \ \\\ \ \ \ +{{\cos }^{2}}\alpha +{{\cos }^{2}}\beta +{{\cos }^{2}}\gamma +2\left( {\cos \alpha \cos \beta +\cos \beta \cos \gamma +\cos \alpha \cos \gamma } \right)=0\\\\\therefore \ \ \left( {{{{\sin }}^{2}}\alpha +{{{\cos }}^{2}}\alpha +{{{\sin }}^{2}}\beta +{{{\cos }}^{2}}\beta +{{{\sin }}^{2}}\gamma {{{\cos }}^{2}}\gamma } \right)\\\ \ \ +2\left( {\cos \alpha \cos \beta +\sin \alpha \sin \beta +\cos \beta \cos \gamma +\sin \beta \sin \gamma +\cos \alpha \cos \gamma +\sin \alpha \sin \gamma } \right)=0\\\\\therefore \ \ 3+2\left[ {\cos \left( {\alpha -\beta } \right)+\cos \left( {\beta -\gamma } \right)+\cos \left( {\gamma -\alpha } \right)} \right]=0\\\\\therefore \ \ 2\left[ {\cos \left( {\alpha -\beta } \right)+\cos \left( {\beta -\gamma } \right)+\cos \left( {\gamma -\alpha } \right)} \right]=-3\\\\\therefore \ \ \cos \left( {\alpha -\beta } \right)+\cos \left( {\beta -\gamma } \right)+\cos \left( {\gamma -\alpha } \right)=-\displaystyle \frac{3}{2}\end{array}$

4.        If $ \displaystyle \sin \alpha +\sin \beta =a$ and $ \displaystyle \cos \alpha +\cos \beta =b$, show that

(a) $ \displaystyle \sin \left( {\alpha +\beta } \right)=\frac{{2ab}}{{{{a}^{2}}+{{b}^{2}}}} \ \ $

(b) $ \displaystyle \cos \left( {\alpha +\beta } \right)=\frac{{{{b}^{2}}-{{a}^{2}}}}{{{{b}^{2}}+{{a}^{2}}}}$

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$ \displaystyle \begin{array}{l}\text{(a)}\ \\\ \ \ \ \sin \alpha +\sin \beta =a\\\\\ \ \ \ \cos \alpha +\cos \beta =b\\\\\therefore \ \ ab=\ \left( {\sin \alpha +\sin \beta } \right)\left( {\cos \alpha +\cos \beta } \right)\\\\\ \ \ \ 2ab=2\sin \alpha \cos \alpha +2\sin \alpha \cos \beta +2\cos \alpha \sin \beta +2\sin \beta \cos \beta \\\\\ \ \ \ 2ab=2\left( {\sin \alpha \cos \beta +\cos \alpha \sin \beta } \right)+2\sin \alpha \cos \alpha +2\sin \beta \cos \beta \\\\\ \ \ \ 2ab=2\sin \left( {\alpha +\beta } \right)+\sin 2\alpha +\sin 2\beta \\\\\ \ \ \ 2ab=2\sin \left( {\alpha +\beta } \right)+2\sin \left( {\alpha +\beta } \right)\cos \left( {\alpha -\beta } \right)\\\\\ \ \ \ 2ab=2\sin \left( {\alpha +\beta } \right)\left[ {1+\cos \left( {\alpha -\beta } \right)} \right]\\\\\ \ \ \ {{a}^{2}}+{{b}^{2}}={{\left( {\sin \alpha +\sin \beta } \right)}^{2}}+{{\left( {\cos \alpha +\cos \beta } \right)}^{2}}\\\\\ \ \ \ {{a}^{2}}+{{b}^{2}}={{\sin }^{2}}\alpha +2\sin \alpha \sin \beta +{{\sin }^{2}}\beta +{{\cos }^{2}}\alpha +2\cos \alpha \cos \beta +{{\cos }^{2}}\beta \\\\\ \ \ \ {{a}^{2}}+{{b}^{2}}={{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha +{{\sin }^{2}}\beta +{{\cos }^{2}}\beta +2\cos \alpha \cos \beta +2\sin \alpha \sin \beta \\\\\ \ \ \ {{a}^{2}}+{{b}^{2}}=2+2\left( {\cos \alpha \cos \beta +\sin \alpha \sin \beta } \right)\\\\\ \ \ \ {{a}^{2}}+{{b}^{2}}=2+2\cos \left( {\alpha -\beta } \right)\\\\\ \ \ \ {{a}^{2}}+{{b}^{2}}=2\left[ {1+\cos \left( {\alpha -\beta } \right)} \right]\\\\\ \ \ \ \displaystyle \frac{{2ab}}{{{{a}^{2}}+{{b}^{2}}}}=\displaystyle \frac{{2\sin \left( {\alpha +\beta } \right)\left[ {1+\cos \left( {\alpha -\beta } \right)} \right]}}{{2\left[ {1+\cos \left( {\alpha -\beta } \right)} \right]}}\\\\\therefore \ \ \sin \left( {\alpha +\beta } \right)=\displaystyle \frac{{2ab}}{{{{a}^{2}}+{{b}^{2}}}}\\\\\text{(b)}\ \\\ \ \ \ \ {{b}^{2}}-{{a}^{2}}={{\left( {\cos \alpha +\cos \beta } \right)}^{2}}-{{\left( {\sin \alpha +\sin \beta } \right)}^{2}}\\\\\ \ \ \ \ {{b}^{2}}-{{a}^{2}}={{\cos }^{2}}\alpha +2\cos \alpha \cos \beta +{{\cos }^{2}}\beta -{{\sin }^{2}}\alpha -2\sin \alpha \sin \beta -{{\sin }^{2}}\beta \\\\\ \ \ \ \ {{b}^{2}}-{{a}^{2}}=2\left( {\cos \alpha \cos \beta -\sin \alpha \sin \beta } \right)+{{\cos }^{2}}\alpha -{{\sin }^{2}}\alpha +{{\cos }^{2}}\beta -{{\sin }^{2}}\beta \\\\\ \ \ \ \ {{b}^{2}}-{{a}^{2}}=2\cos \left( {\alpha +\beta } \right)+\cos 2\alpha +\cos 2\beta \\\\\ \ \ \ \ {{b}^{2}}-{{a}^{2}}=2\cos \left( {\alpha +\beta } \right)+2\cos \left( {\alpha +\beta } \right)\cos \left( {\alpha -\beta } \right)\\\\\ \ \ \ \ {{b}^{2}}-{{a}^{2}}=2\cos \left( {\alpha +\beta } \right)\left[ {1+\cos \left( {\alpha -\beta } \right)} \right]\\\\\therefore \ \ \ \displaystyle \frac{{{{b}^{2}}-{{a}^{2}}}}{{{{b}^{2}}+{{a}^{2}}}}=\displaystyle \frac{{2\cos \left( {\alpha +\beta } \right)\left[ {1+\cos \left( {\alpha -\beta } \right)} \right]}}{{2\left[ {1+\cos \left( {\alpha -\beta } \right)} \right]}}\\\\\therefore \ \ \ \cos \left( {\alpha +\beta } \right)=\displaystyle \frac{{{{b}^{2}}-{{a}^{2}}}}{{{{b}^{2}}+{{a}^{2}}}}\end{array}$ $ \displaystyle \begin{array}{l}\underline{{\text{Alternative Method}}}\\\\\ \ \ \ \sin \alpha +\sin \beta =a\\\\\ \ \ \ 2\sin \displaystyle \frac{{a+\beta }}{2}\cos \displaystyle \frac{{a+\beta }}{2}=a\\\\\ \ \ \ \cos \alpha +\cos \beta =b\\\\\ \ \ \ 2\cos \displaystyle \frac{{a+\beta }}{2}\cos \displaystyle \frac{{a+\beta }}{2}=b\\\\\therefore \ \ \displaystyle \frac{{2\sin \displaystyle \frac{{a+\beta }}{2}\cos \displaystyle \frac{{a+\beta }}{2}}}{{2\cos \displaystyle \frac{{a+\beta }}{2}\cos \displaystyle \frac{{a+\beta }}{2}}}=\displaystyle \frac{a}{b}\\\\\therefore \ \ \tan \displaystyle \frac{{a+\beta }}{2}\\\\\ \ \ \ \text{Let}\ \text{ }a+\beta =\theta ,\ \text{then}\ \tan \displaystyle \frac{{a+\beta }}{2}=\tan \displaystyle \frac{\theta }{2}\ \\\\\ \ \ \ \text{Let}\ \tan \displaystyle \frac{\theta }{2}=t,\ \text{then}\ t=\displaystyle \frac{a}{b}\\\\\text{(a)}\sin \left( {a+\beta } \right)=\sin \theta \\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{2t}}{{1+{{t}^{2}}}}\ \ \left[ {\text{t - formulae}} \right]\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{2\left( {\displaystyle \frac{a}{b}} \right)}}{{1+{{{\left( {\displaystyle \frac{a}{b}} \right)}}^{2}}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{\displaystyle \frac{{2a}}{b}}}{{\displaystyle \frac{{{{a}^{2}}+{{b}^{2}}}}{{{{b}^{2}}}}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{2ab}}{{{{a}^{2}}+{{b}^{2}}}}\\\\\text{(b)}\ \cos \left( {a+\beta } \right)=\cos \theta \\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{1-{{t}^{2}}}}{{1+{{t}^{2}}}}\ \ \left[ {\text{t - formulae}} \right]\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{1-{{{\left( {\displaystyle \frac{a}{b}} \right)}}^{2}}}}{{1+{{{\left( {\displaystyle \frac{a}{b}} \right)}}^{2}}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{\displaystyle \frac{{{{b}^{2}}-{{a}^{2}}}}{{{{b}^{2}}}}}}{{\displaystyle \frac{{{{b}^{2}}+{{a}^{2}}}}{{{{b}^{2}}}}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{{{b}^{2}}-{{a}^{2}}}}{{{{b}^{2}}+{{a}^{2}}}}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \end{array}$

5.        Without using tables, prove that

(a) $ \displaystyle \text{sin 47}{}^\circ +\cos 77{}^\circ =\cos 17{}^\circ $

(b) $ \displaystyle \cos 80{}^\circ +\cos 40{}^\circ -\cos 20{}^\circ =0$

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$ \displaystyle \begin{array}{l}\text{(a)}\\\ \ \ \ \text{sin 47}{}^\circ +\cos 77{}^\circ \\\\=\text{sin 47}{}^\circ +\cos \left( {90{}^\circ -13{}^\circ } \right)\\\\=\text{sin 47}{}^\circ +\sin 13{}^\circ \\\\=2\sin \displaystyle \frac{{\text{47}{}^\circ +13{}^\circ }}{2}\cos \displaystyle \frac{{\text{47}{}^\circ -13{}^\circ }}{2}\\\\=2\sin \text{30}{}^\circ \cos \text{17}{}^\circ \\\\=2\left( {\displaystyle \frac{1}{2}} \right)\cos \text{17}{}^\circ \\\\=\cos \text{17}{}^\circ \\\\\text{(b)}\\\ \ \ \cos 80{}^\circ +\cos 40{}^\circ -\cos 20{}^\circ \\\\=2\cos \displaystyle \frac{{\text{80}{}^\circ +40{}^\circ }}{2}\cos \displaystyle \frac{{\text{80}{}^\circ -40{}^\circ }}{2}-\cos 20{}^\circ \\\\=2\cos \text{60}{}^\circ \cos 2\text{0}{}^\circ -\cos 20{}^\circ \\\\=2\left( {\displaystyle \frac{1}{2}} \right)\cos 2\text{0}{}^\circ -\cos 20{}^\circ \\\\=\cos 2\text{0}{}^\circ -\cos 20{}^\circ \\\\=0\end{array}$

6.        Prove that $ \displaystyle {{(\sin \alpha -\sin \beta )}^{2}}+{{(\cos \alpha -\cos \beta )}^{2}}=4{{\sin }^{2}}\left( {\frac{{\alpha -\beta }}{2}} \right)$.

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$ \displaystyle \begin{array}{l}\ \ \ \ {{\left( {\sin \alpha -\sin \beta } \right)}^{2}}+{{\left( {\cos \alpha -\cos \beta } \right)}^{2}}\\\\=\ \ {{\cos }^{2}}\alpha -2\cos \alpha \cos \beta +{{\cos }^{2}}\beta \\\ \ \ \ +\ {{\sin }^{2}}\alpha -2\sin \alpha \sin \beta +{{\sin }^{2}}\beta \\\\=\ \ {{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha +{{\sin }^{2}}\beta +{{\cos }^{2}}\beta \\\ \ \ \ -2\cos \alpha \cos \beta -2\sin \alpha \sin \beta \\\\=\ \ 2-2\left( {\cos \alpha \cos \beta +\sin \alpha \sin \beta } \right)\\\\=\ \ 2\left[ {1-\cos \left( {\alpha -\beta } \right)} \right]\\\\=\ \ 2\left[ {1-\cos 2\left( {\displaystyle \frac{{\alpha -\beta }}{2}} \right)} \right]\\\\=\ \ 2\left[ {1-\left( {1-2{{{\sin }}^{2}}\left( {\displaystyle \frac{{\alpha -\beta }}{2}} \right)} \right)} \right]\\\\=\ \ 2\left[ {2{{{\sin }}^{2}}\left( {\displaystyle \frac{{\alpha -\beta }}{2}} \right)} \right]\\\\=4{{\sin }^{2}}\left( {\displaystyle \frac{{\alpha -\beta }}{2}} \right)\end{array}$

7.        Prove that $ \displaystyle {{(\sin \alpha +\sin \beta )}^{2}}+{{(\cos \alpha +\cos \beta )}^{2}}=4{{\cos }^{2}}\left( {\frac{{\alpha -\beta }}{2}} \right)$.

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$ \displaystyle \begin{array}{l}\ \ \ \ {{\left( {\sin \alpha +\sin \beta } \right)}^{2}}+{{\left( {\cos \alpha +\cos \beta } \right)}^{2}}\\\\=\ \ {{\cos }^{2}}\alpha +2\cos \alpha \cos \beta +{{\cos }^{2}}\beta \\\ \ \ \ +\ {{\sin }^{2}}\alpha +2\sin \alpha \sin \beta +{{\sin }^{2}}\beta \\\\=\ \ {{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha +{{\sin }^{2}}\beta +{{\cos }^{2}}\beta \\\ \ \ \ +2\cos \alpha \cos \beta +2\sin \alpha \sin \beta \\\\=\ \ 2+2\left( {\cos \alpha \cos \beta +\sin \alpha \sin \beta } \right)\\\\=\ \ 2\left[ {1+\cos \left( {\alpha -\beta } \right)} \right]\\\\=\ \ 2\left[ {1+\cos 2\left( {\displaystyle \frac{{\alpha -\beta }}{2}} \right)} \right]\\\\=\ \ 2\left[ {1+\left( {2{{{\cos }}^{2}}\left( {\displaystyle \frac{{\alpha -\beta }}{2}} \right)-1} \right)} \right]\\\\=\ \ 2\left[ {2{{{\cos }}^{2}}\left( {\displaystyle \frac{{\alpha -\beta }}{2}} \right)} \right]\\\\=4{{\cos }^{2}}\left( {\displaystyle \frac{{\alpha -\beta }}{2}} \right)\end{array}$

8.        Prove that $ \displaystyle \frac{{\sin \theta +\sin 2\theta +\sin 4\theta +\sin 5\theta }}{{\cos \theta +\cos 2\theta +\cos 4\theta +\cos 5\theta }}=\tan 3\theta $

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$ \displaystyle \begin{array}{l}\ \ \ \displaystyle \frac{{\sin \theta +\sin 2\theta +\sin 4\theta +\sin 5\theta }}{{\cos \theta +\cos 2\theta +\cos 4\theta +\cos 5\theta }}\\\\=\ \displaystyle \frac{{\sin 5\theta +\sin \theta +\sin 4\theta +\sin 2\theta }}{{\cos 5\theta +\cos \theta +\cos 4\theta +\cos 2\theta }}\\\\=\ \displaystyle \frac{{2\sin \displaystyle \frac{{5\theta +\theta }}{2}\cos \displaystyle \frac{{5\theta -\theta }}{2}+2\sin \displaystyle \frac{{4\theta +2\theta }}{2}\cos \displaystyle \frac{{4\theta -2\theta }}{2}}}{{2\cos \displaystyle \frac{{5\theta +\theta }}{2}\cos \displaystyle \frac{{5\theta -\theta }}{2}+2\cos \displaystyle \frac{{4\theta +2\theta }}{2}\cos \displaystyle \frac{{4\theta -2\theta }}{2}}}\\\\=\ \displaystyle \frac{{2\left( {\sin 3\theta \cos 2\theta +\sin 3\theta \cos \theta } \right)}}{{2\left( {\cos 3\theta \cos 2\theta +\cos 3\theta \cos \theta } \right)}}\\\\=\ \displaystyle \frac{{\sin 3\theta \left( {\cos 2\theta +\cos \theta } \right)}}{{\cos 3\theta \left( {\cos 2\theta +\cos \theta } \right)}}\\\\=\tan 3\theta \end{array}$

9.        If $\displaystyle \tan \alpha =\frac{m}{{m+1}}$ and $ \displaystyle \tan \beta =\frac{1}{{2m+1}}$, find the acute angle $ \displaystyle \theta$ such that $ \displaystyle \cot \theta =\tan \left( {\alpha +\beta } \right)$.

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$ \displaystyle \begin{array}{l}\ \ \ \ \tan \alpha =\displaystyle \frac{m}{{m+1}},\tan \beta =\displaystyle \frac{1}{{2m+1}}\\\\\ \ \ \ \cot \theta =\tan \left( {\alpha +\beta } \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{\tan \alpha +\tan \beta }}{{1-\tan \alpha \tan \beta }}\\\\\ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{\displaystyle \frac{m}{{m+1}}+\displaystyle \frac{1}{{2m+1}}}}{{1-\displaystyle \frac{m}{{m+1}}\times \displaystyle \frac{1}{{2m+1}}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{\displaystyle \frac{{2{{m}^{2}}+m+m+1}}{{\left( {m+1} \right)\left( {2m+1} \right)}}}}{{\displaystyle \frac{{2{{m}^{2}}+3m+1-m}}{{\left( {m+1} \right)\left( {2m+1} \right)}}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{2{{m}^{2}}+2m+1}}{{2{{m}^{2}}+2m+1}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ =1\\\\\therefore \ \ \cot \theta =\cot 45{}^\circ \\\\\therefore \ \ \theta =45{}^\circ \end{array}$

10.       Prove that

(a) $ \displaystyle \frac{{1+\sin 2\alpha -\cos 2\alpha }}{{1+\sin 2\alpha +\cos 2\alpha }}=\tan \alpha $

(b) $ \displaystyle \frac{{1+\sin 2\alpha +\cos 2\alpha }}{{1+\sin 2\alpha -\cos 2\alpha }}=\cot \alpha $

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$ \displaystyle \begin{array}{l}\text{(a)}\\\ \ \ \displaystyle \frac{{1+\sin 2\alpha -\cos 2\alpha }}{{1+\sin 2\alpha +\cos 2\alpha }}\\\\=\displaystyle \frac{{1+2\sin \alpha \cos \alpha -1+2{{{\sin }}^{2}}\alpha }}{{1+2\sin \alpha \cos \alpha +2{{{\cos }}^{2}}\alpha -1}}\\\\=\displaystyle \frac{{2\sin \alpha \left( {\sin \alpha +\cos \alpha } \right)}}{{2\cos \alpha \left( {\sin \alpha +\cos \alpha } \right)}}\\\\=\tan \alpha \\\\\text{(b)}\\\ \ \ \displaystyle \frac{{1+\sin 2\alpha +\cos 2\alpha }}{{1+\sin 2\alpha -\cos 2\alpha }}\\\\=\displaystyle \frac{{1+2\sin \alpha \cos \alpha +2{{{\cos }}^{2}}\alpha -1}}{{1+2\sin \alpha \cos \alpha -1+2{{{\sin }}^{2}}\alpha }}\\\\=\displaystyle \frac{{2\cos \alpha \left( {\sin \alpha +\cos \alpha } \right)}}{{2\sin \alpha \left( {\sin \alpha +\cos \alpha } \right)}}\\\\=\cot \alpha \end{array}$

11.       If $ \displaystyle \sec \theta + \tan \theta = 3$, where $ \displaystyle \theta$ lies in the first quadrant, then find the value of $ \displaystyle \cos \theta$.

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$ \displaystyle \begin{array}{l}\ \ \ \sec \theta +\tan \theta =3\ ---(1)\\\\\ \ \ \displaystyle \frac{{\left( {\sec \theta +\tan \theta } \right)\left( {\sec \theta -\tan \theta } \right)}}{{\sec \theta -\tan \theta }}=3\\\\\ \ \ \displaystyle \frac{{{{{\sec }}^{2}}\theta -{{{\tan }}^{2}}\theta }}{{\sec \theta -\tan \theta }}=3\\\\\ \ \ \displaystyle \frac{1}{{\sec \theta -\tan \theta }}=3\\\\\therefore \ \ \sec \theta -\tan \theta =\displaystyle \frac{1}{3}\ ---(2)\\\\\ \ \ (1)+(2)\Rightarrow 2\sec \theta =\displaystyle \frac{{10}}{3}\\\\\therefore \ \sec \theta =\displaystyle \frac{5}{3}\\\\\therefore \ \cos \theta =\displaystyle \frac{3}{5}\end{array}$

12.       If $ \displaystyle \operatorname{cosec}\theta -\cot \theta =\frac{1}{5}$, where $ \displaystyle \theta$ lies in the first quadrant, then find the value of $ \displaystyle \sin \theta$.

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$ \displaystyle \begin{array}{l}\ \ \ \ \operatorname{cosec}\theta -\cot \theta =\displaystyle \frac{1}{5}\ \ \ ---(1)\\\\\therefore \ \ \displaystyle \frac{{\left( {\operatorname{cosec}\theta -\cot \theta } \right)\left( {\operatorname{cosec}\theta +\cot \theta } \right)}}{{\left( {\operatorname{cosec}\theta +\cot \theta } \right)}}=\displaystyle \frac{1}{5}\\\\\therefore \ \ \displaystyle \frac{{{{{\operatorname{cosec}}}^{2}}\theta -{{{\cot }}^{2}}\theta }}{{\operatorname{cosec}\theta +\cot \theta }}=\displaystyle \frac{1}{5}\\\\\therefore \ \ \displaystyle \frac{1}{{\operatorname{cosec}\theta +\cot \theta }}=\displaystyle \frac{1}{5}\\\ \ \ \left[ {\because 1+{{{\cot }}^{2}}\theta ={{{\operatorname{cosec}}}^{2}}\theta } \right]\\\\\therefore \ \ \operatorname{cosec}\theta +\cot \theta =5\ \ \ ---(2)\\\\\ \ \ (1)+(2)\Rightarrow 2\operatorname{cosec}\theta =\displaystyle \frac{{26}}{5}\\\\\therefore \ \operatorname{cosec}\theta =\displaystyle \frac{{13}}{5}\\\\\therefore \ \sin \theta =\displaystyle \frac{5}{{13}}\end{array}$

13.       If $ \displaystyle \cos \theta +\sin \theta =\sqrt{2}\cos \theta $, then prove that $ \displaystyle \cos \theta -\sin \theta =\sqrt{2}\sin \theta $.

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$ \displaystyle \begin{array}{l}\ \ \ \ \cos \theta +\sin \theta =\sqrt{2}\cos \theta \\\\\therefore \ \ {{\left( {\cos \theta +\sin \theta } \right)}^{2}}=2{{\cos }^{2}}\theta \ \\\\\ \ \ \ {{\cos }^{2}}\theta +2\sin \theta \cos \theta +{{\sin }^{2}}\theta =2{{\cos }^{2}}\theta \ \\\\\therefore \ \ {{\cos }^{2}}\theta -2\sin \theta \cos \theta -{{\sin }^{2}}\theta =0\\\\\therefore \ \ {{\cos }^{2}}\theta -2\sin \theta \cos \theta -{{\sin }^{2}}\theta +2{{\sin }^{2}}\theta =2{{\sin }^{2}}\theta \\\\\therefore \ \ {{\cos }^{2}}\theta -2\sin \theta \cos \theta +{{\sin }^{2}}\theta =2{{\sin }^{2}}\theta \\\\\therefore \ \ {{\left( {\cos \theta -\sin \theta } \right)}^{2}}=2{{\sin }^{2}}\theta \\\\\therefore \ \ \cos \theta -\sin \theta =\sqrt{2}\sin \theta \end{array}$

14.       If $ \displaystyle \sin \left( {\alpha -\beta } \right)=\cos \left( {\alpha +\beta } \right)=\frac{1}{2}$, find the positive acute angles $ \displaystyle \alpha$ and $ \displaystyle \beta$.

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$ \displaystyle \begin{array}{l}\ \ \ \sin \left( {\alpha -\beta } \right)=\displaystyle \frac{1}{2}\\\\\therefore \ \alpha -\beta =30{}^\circ \,\ ---(1)\\\\\ \ \ \cos \left( {\alpha +\beta } \right)=\displaystyle \frac{1}{2}\\\\\therefore \ \alpha +\beta =60{}^\circ \ \ ---(2)\\\\\ \ \ (1)+(2)\Rightarrow 2\alpha =90{}^\circ \\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \alpha =45{}^\circ \\\\\ \ \ (2)-(1)\Rightarrow 2\beta =30{}^\circ \\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \beta =15{}^\circ \end{array}$

15.       Prove that $ \displaystyle \frac{{2\sin \theta }}{{1+\sin \theta +\cos \theta }}+\frac{{\cos \theta }}{{1+\sin \theta }}=1$.

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$ \displaystyle \begin{array}{l}\ \ \ \ \displaystyle \frac{{2\sin \theta }}{{1+\sin \theta +\cos \theta }}+\displaystyle \frac{{\cos \theta }}{{1+\sin \theta }}\\\\=\ \ \displaystyle \frac{{2\sin \theta \left( {1+\sin \theta } \right)+\cos \theta \left( {1+\sin \theta +\cos \theta } \right)}}{{\left( {1+\sin \theta +\cos \theta } \right)\left( {1+\sin \theta } \right)}}\\\\=\ \ \displaystyle \frac{{2\sin \theta +2{{{\sin }}^{2}}\theta +\cos \theta +\sin \theta \cos \theta +{{{\cos }}^{2}}\theta }}{{\left( {1+\sin \theta +\cos \theta } \right)\left( {1+\sin \theta } \right)}}\\\\=\ \ \displaystyle \frac{{{{{\sin }}^{2}}\theta +{{{\cos }}^{2}}\theta +\sin \theta +\sin \theta +{{{\sin }}^{2}}\theta +\cos \theta +\sin \theta \cos \theta }}{{\left( {1+\sin \theta +\cos \theta } \right)\left( {1+\sin \theta } \right)}}\\\\=\ \ \displaystyle \frac{{\left( {1+\sin \theta } \right)+\sin \theta \left( {1+\sin \theta } \right)+\cos \theta \left( {1+\sin \theta } \right)}}{{\left( {1+\sin \theta +\cos \theta } \right)\left( {1+\sin \theta } \right)}}\\\\=\ \ \displaystyle \frac{{\left( {1+\sin \theta } \right)\left( {1+\sin \theta +\cos \theta } \right)}}{{\left( {1+\sin \theta +\cos \theta } \right)\left( {1+\sin \theta } \right)}}\\\\=1\end{array}$

16.      If $ \displaystyle \tan \theta =\frac{4}{5}$, where $ \displaystyle \theta$ lies in the first quadrant, find the exact value of $ \displaystyle \frac{{5\sin \theta -3\cos \theta }}{{\sin \theta +2\cos \theta }}$.

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$ \displaystyle \begin{array}{l}\ \ \ \ \tan \theta =\displaystyle \frac{4}{5},\\\\\ \ \ \ \displaystyle \frac{{5\sin \theta -3\cos \theta }}{{\sin \theta +2\cos \theta }}\\\\=\ \displaystyle \frac{{5\displaystyle \frac{{\sin \theta }}{{\cos \theta }}-3\displaystyle \frac{{\cos \theta }}{{\cos \theta }}}}{{\displaystyle \frac{{\sin \theta }}{{\cos \theta }}+2\displaystyle \frac{{\cos \theta }}{{\cos \theta }}}}\\\\=\ \displaystyle \frac{{5\tan \theta -3}}{{\tan \theta +2}}\\\\=\ \displaystyle \frac{{5\left( {\displaystyle \frac{4}{5}} \right)-3}}{{\left( {\displaystyle \frac{4}{5}} \right)+2}}\\\\=\displaystyle \frac{1}{{\left( {\displaystyle \frac{{14}}{5}} \right)}}\\\\=\displaystyle \frac{5}{{14}}\end{array}$

17.       If $ \displaystyle \tan \theta +\sin \theta =m$ and $ \displaystyle \tan \theta -\sin \theta =n$, show that $ \displaystyle {{m}^{2}}-{{n}^{2}}=4\sqrt{{mn}}$.

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$ \displaystyle \begin{array}{l}\ \ \ \tan \theta +\sin \theta =m,\ \\\\\ \ \ \tan \theta -\sin \theta =n\\\\\therefore \ \ m+n=2\tan \theta \\\\\therefore \ \ m-n=2\sin \theta \\\\\therefore \ \ \left( {m+n} \right)\left( {m-n} \right)=2\tan \theta \times 2\sin \theta \\\\\therefore \ \ {{m}^{2}}-{{n}^{2}}=4\sin \theta \tan \theta \\\\\ \ \ \ mn=\left( {\tan \theta +\sin \theta } \right)\left( {\tan \theta -\sin \theta } \right)\\\\\ \ \ \ mn={{\tan }^{2}}\theta -{{\sin }^{2}}\theta \\\\\ \ \ \ mn=\displaystyle \frac{{{{{\sin }}^{2}}\theta }}{{{{{\cos }}^{2}}\theta }}-{{\sin }^{2}}\theta \\\\\ \ \ \ mn=\displaystyle \frac{{{{{\sin }}^{2}}\theta -{{{\sin }}^{2}}\theta {{{\cos }}^{2}}\theta }}{{{{{\cos }}^{2}}\theta }}\\\\\ \ \ \ mn=\displaystyle \frac{{{{{\sin }}^{2}}\theta \left( {1-{{{\cos }}^{2}}\theta } \right)}}{{{{{\cos }}^{2}}\theta }}\\\\\ \ \ \ mn=\displaystyle \frac{{{{{\sin }}^{2}}\theta {{{\sin }}^{2}}\theta }}{{{{{\cos }}^{2}}\theta }}\\\\\ \ \ \ mn={{\sin }^{2}}\theta {{\tan }^{2}}\theta \\\\\therefore \ \ \sqrt{{mn}}=\sin \theta \tan \theta \\\\\therefore \ \ 4\sqrt{{mn}}=4\sin \theta \tan \theta \\\\\therefore \ \ {{m}^{2}}-{{n}^{2}}=4\sqrt{{mn}}\end{array}$

18.       Without using tables, prove that $ \displaystyle \cos 105{}^\circ +\cos 15{}^\circ =\sin 75{}^\circ -\sin 15{}^\circ $.

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$ \displaystyle \begin{array}{l}\ \ \ \ \cos 105{}^\circ +\cos 15{}^\circ \\\\=\,\ 2\cos \displaystyle \frac{{105{}^\circ +15{}^\circ }}{2}\cos \displaystyle \frac{{105{}^\circ -15{}^\circ }}{2}\\\\=\,\ 2\cos 60{}^\circ \cos 45{}^\circ \\\\=\,\ 2\left( {\displaystyle \frac{1}{2}} \right)\displaystyle \frac{{\sqrt{2}}}{2}\\\\=\,\ \displaystyle \frac{{\sqrt{2}}}{2}\\\\\ \ \ \ \sin 75{}^\circ -\sin 15{}^\circ \\\\=\,\ 2\cos \displaystyle \frac{{75{}^\circ +15{}^\circ }}{2}\sin \displaystyle \frac{{75{}^\circ -15{}^\circ }}{2}\\\\=\,\ 2\cos 45{}^\circ \sin 30{}^\circ \\\\=\,\ 2\left( {\displaystyle \frac{{\sqrt{2}}}{2}} \right)\displaystyle \frac{1}{2}\\\\=\,\ \displaystyle \frac{{\sqrt{2}}}{2}\\\\\therefore \ \cos 105{}^\circ +\cos 15{}^\circ =\sin 75{}^\circ -\sin 15{}^\circ \end{array}$

19.       Prove that $ \displaystyle \sin \theta +\cos \theta =\sqrt{2}\sin \left( {\theta +45{}^\circ } \right)=\sqrt{2}\cos \left( {\theta -45{}^\circ } \right)$.

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$ \displaystyle \begin{array}{l}\ \ \ \ \sin \theta +\cos \theta \\\\=\,\sqrt{2}\ \left( {\displaystyle \frac{1}{{\sqrt{2}}}\sin \theta +\displaystyle \frac{1}{{\sqrt{2}}}\cos \theta } \right)\\\\=\,\sqrt{2}\ \left( {\sin \theta \cos 45{}^\circ +\cos \theta \sin 45{}^\circ } \right)\\\ \ \ \ \left[ {\because \sin 45{}^\circ =\cos 45{}^\circ =\displaystyle \frac{1}{{\sqrt{2}}}} \right]\\\\=\,\sqrt{2}\ \sin \left( {\theta +45{}^\circ } \right)\\\\\ \ \ \ \sin \theta +\cos \theta \\\\=\,\sqrt{2}\ \left( {\displaystyle \frac{1}{{\sqrt{2}}}\sin \theta +\displaystyle \frac{1}{{\sqrt{2}}}\cos \theta } \right)\\\\=\,\sqrt{2}\ \left( {\sin \theta \sin 45{}^\circ +\cos \theta \cos 45{}^\circ } \right)\\\ \ \ \ \left[ {\because \sin 45{}^\circ =\cos 45{}^\circ =\displaystyle \frac{1}{{\sqrt{2}}}} \right]\\\\=\,\sqrt{2}\ \left( {\cos \theta \cos 45{}^\circ +\sin \theta \sin 45{}^\circ } \right)\\\\=\,\sqrt{2}\ \cos \left( {\theta -45{}^\circ } \right)\end{array}$

20.       In $ \displaystyle \vartriangle ABC$, if $ \displaystyle \cot A+\cot B+\cot C=\sqrt{3}$, prove that $ \displaystyle \vartriangle ABC$ is an equilateral triangle.

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$ \displaystyle \begin{array}{l}\ \ \ \ \text{In}\ \vartriangle ABC,\\\\\ \ \ \ A+B+C=180{}^\circ \\\\\ \ \ \ A+B=180{}^\circ -C\\\\\ \ \ \ \tan \left( {A+B} \right)=\tan \left( {180{}^\circ -C} \right)\\\\\ \ \ \ \displaystyle \frac{{\tan A+\tan B}}{{1-\tan A\tan B}}=-\tan C\\\\\ \ \ \ \displaystyle \frac{{\displaystyle \frac{1}{{\cot A}}+\displaystyle \frac{1}{{\cot B}}}}{{1-\displaystyle \frac{1}{{\cot A\cot B}}}}=-\displaystyle \frac{1}{{\cot C}}\\\\\ \ \ \ \displaystyle \frac{{\displaystyle \frac{{\cot A+\cot B}}{{\cot A\cot B}}}}{{\displaystyle \frac{{\cot A\cot B-1}}{{\cot A\cot B}}}}=-\displaystyle \frac{1}{{\cot C}}\\\\\ \ \ \ \displaystyle \frac{{\cot A+\cot B}}{{\cot A\cot B-1}}=-\displaystyle \frac{1}{{\cot C}}\\\\\therefore \ \ \cot A\cot C+\cot B\cot C=1-\cot A\cot B\\\\\therefore \ \ \cot A\cot B+\cot B\cot C+\cot A\cot C=1\ \\\\\ \ \ \ \text{Let}\ \cot A=x,\ \cot B=y,\ \cot C=z\\\\\therefore \ \ xy+yz+xz=1\ ---(1)\\\\\ \ \ \ x+y+z=\sqrt{3}\,\ \left[ {\because given} \right]\\\\\therefore \ \ {{\left( {x+y+z} \right)}^{2}}=3\\\\\ \ \ {{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2\left( {xy+yz+xz} \right)=3\\\\\therefore \ \ {{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2(1)=3\\\\\therefore \ \ {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-1=0\\\\\therefore \ \ {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-xz=0\ ---(2)\\\\\therefore \ \ 2{{x}^{2}}+2{{y}^{2}}+2{{z}^{2}}-2xy-2yz-2xz=0\\\\\therefore \ \left( {{{x}^{2}}-2xy+{{y}^{2}}} \right)+\left( {{{y}^{2}}-2yz+{{z}^{2}}} \right)+\left( {{{x}^{2}}-2xz+{{z}^{2}}} \right)=0\\\\\therefore \ {{\left( {x-y} \right)}^{2}}+{{\left( {y-z} \right)}^{2}}+{{\left( {x-z} \right)}^{2}}=0\\\\\therefore \ \ x-y=0\Rightarrow x=y\\\\\ \ \ y-z=0\Rightarrow y=z\\\\\ \ \ x-z=0\Rightarrow x=z\\\\\therefore \ \ x=y=z\Rightarrow \cot A=\cot B=\cot C\\\\\therefore \ \ A=B=C\\\\\therefore \ \ \vartriangle ABC\ \text{is an equilateral triangle}\text{.}\ \end{array}$

Exercise (11.4) No (1 to 26) - Solution

1.        Use the compound angle formulae to find the following in surd form:

(a) sin 105°	(b) cos15°	(c) tan 75°
(d) cos 165°	(e) sin 345°	(f) sin 75°
(g) cos105°	(h) tan 165°	(i) cos 345°
(j) tan 15°	(k) sin (–15°)	(l) sec (–75°)

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$ \displaystyle \begin{array}{l} \text{(a)}\ \ \ \ \ \sin 105{}^\circ \\\\\ \ \ \ \ \ =\sin (180{}^\circ -75{}^\circ )\\\\\ \ \ \ \ \ =\sin 75{}^\circ \\\\\ \ \ \ \ \ =\sin (30{}^\circ +45{}^\circ )\\\\\ \ \ \ \ \ =\sin 30{}^\circ \cos 45{}^\circ +\cos 30{}^\circ \sin 45{}^\circ \\\\\ \ \ \ \ \ =\displaystyle \frac{1}{2}\times \displaystyle \frac{{\sqrt{2}}}{2}+\displaystyle \frac{{\sqrt{3}}}{2}\times \displaystyle \frac{{\sqrt{2}}}{2}\\\\\ \ \ \ \ \ =\displaystyle \frac{{\sqrt{2}+\sqrt{6}}}{4}\\\\ \text{(b)}\ \ \ \ \ \cos 15{}^\circ \\\\\ \ \ \ \ \ =\cos (45{}^\circ -30{}^\circ )\\\\\ \ \ \ \ \ =\cos 45{}^\circ \cos 30{}^\circ +\sin 45{}^\circ \sin 30{}^\circ \\\\\ \ \ \ \ \ =\displaystyle \frac{{\sqrt{2}}}{2}\times \displaystyle \frac{{\sqrt{3}}}{2}+\displaystyle \frac{{\sqrt{2}}}{2}\times \displaystyle \frac{1}{2}\\\\\ \ \ \ \ \ =\displaystyle \frac{{\sqrt{6}+\sqrt{2}}}{4}\\\\\end{array}$

$ \displaystyle \begin{array}{l} \text{(c)}\ \ \ \ \ \tan 75{}^\circ \\\\\ \ \ \ \ \ =\tan (45{}^\circ +30{}^\circ )\\\\\ \ \ \ \ \ =\displaystyle \frac{{\tan 45{}^\circ +\tan 30{}^\circ }}{{1-\tan 45{}^\circ \tan 30{}^\circ }}\\\\\ \ \ \ \ \ =\displaystyle \frac{{1+\displaystyle \frac{1}{{\sqrt{3}}}}}{{1-\displaystyle \frac{1}{{\sqrt{3}}}}}\\\\\ \ \ \ \ \ =\displaystyle \frac{{\sqrt{3}+1}}{{\sqrt{3}-1}}\times \displaystyle \frac{{\sqrt{3}+1}}{{\sqrt{3}+1}}\\\\\ \ \ \ \ \ =\displaystyle \frac{{3+2\sqrt{3}+1}}{2}\\\\\ \ \ \ \ \ =2+\sqrt{3}\\\\ \text{(d)}\ \ \ \ \ \cos 165{}^\circ \\\\\ \ \ \ \ \ =\cos (180{}^\circ -15{}^\circ )\\\\\ \ \ \ \ \ =-\cos (180{}^\circ -15{}^\circ )\\\\\ \ \ \ \ \ =-\left( {\cos 45{}^\circ \cos 30{}^\circ +\sin 45{}^\circ \sin 30{}^\circ } \right)\\\\\ \ \ \ \ \ =-\left( {\displaystyle \frac{{\sqrt{2}}}{2}\times \displaystyle \frac{{\sqrt{3}}}{2}+\displaystyle \frac{{\sqrt{2}}}{2}\times \displaystyle \frac{1}{2}} \right)\\\\\ \ \ \ \ \ =\displaystyle \frac{{-\sqrt{6}-\sqrt{2}}}{4}\\\\\end{array}$

$ \displaystyle \begin{array}{l} \text{(e)}\ \ \ \ \ \sin 345{}^\circ \\\\\ \ \ \ \ \ =\sin (360{}^\circ -15{}^\circ )\\\\\ \ \ \ \ \ =-\sin 15{}^\circ \\\\\ \ \ \ \ \ =-\sin (45{}^\circ -30{}^\circ )\\\\\ \ \ \ \ \ =-(\sin 45{}^\circ \cos 30{}^\circ -\cos 45{}^\circ \sin 30{}^\circ )\\\\\ \ \ \ \ \ =-\left( {\displaystyle \frac{{\sqrt{2}}}{2}\times \displaystyle \frac{{\sqrt{3}}}{2}-\displaystyle \frac{{\sqrt{2}}}{2}\times \displaystyle \frac{1}{2}} \right)\\\\\ \ \ \ \ \ =-\left( {\displaystyle \frac{{\sqrt{6}-\sqrt{2}}}{4}} \right)\\\\\ \ \ \ \ \ =\displaystyle \frac{{\sqrt{2}-\sqrt{6}}}{4}\\\\ \text{(f)}\ \ \ \ \ \sin 75{}^\circ \\\\\ \ \ \ \ \ =\sin (45{}^\circ +30{}^\circ )\\\\\ \ \ \ \ \ =\sin 45{}^\circ \cos 30{}^\circ +\cos 45{}^\circ \sin 30{}^\circ \\\\\ \ \ \ \ \ =\displaystyle \frac{{\sqrt{2}}}{2}\times \displaystyle \frac{{\sqrt{3}}}{2}+\displaystyle \frac{{\sqrt{2}}}{2}\times \displaystyle \frac{1}{2}\\\\\ \ \ \ \ \ =\displaystyle \frac{{\sqrt{6}+\sqrt{2}}}{4}\\\\\end{array}$

$ \displaystyle \begin{array}{l} \text{(g)}\ \ \ \ \ \cos 105{}^\circ \\\\\ \ \ \ \ \ =\cos (180{}^\circ -75{}^\circ )\\\\\ \ \ \ \ \ =-\cos 75{}^\circ \\\\\ \ \ \ \ \ =-\cos (30{}^\circ +45{}^\circ )\\\\\ \ \ \ \ \ =-\left( {\cos 30{}^\circ \cos 45{}^\circ -\sin 30{}^\circ \sin 45{}^\circ } \right)\\\\\ \ \ \ \ \ =-\left( {\displaystyle \frac{{\sqrt{3}}}{2}\times \displaystyle \frac{{\sqrt{2}}}{2}-\displaystyle \frac{1}{2}\times \displaystyle \frac{{\sqrt{2}}}{2}} \right)\\\\\ \ \ \ \ \ =\displaystyle \frac{{\sqrt{2}-\sqrt{6}}}{4}\\\\ \text{(h)}\ \ \ \ \ \tan 165{}^\circ \\\\\ \ \ \ \ \ =\tan (180{}^\circ -15{}^\circ )\\\\\ \ \ \ \ \ =-\tan 15{}^\circ \\\\\ \ \ \ \ \ =-\tan (45{}^\circ -30{}^\circ )\\\\\ \ \ \ \ \ =-\left( {\displaystyle \frac{{\tan 45{}^\circ -\tan 30{}^\circ }}{{1+\tan 45{}^\circ \tan 30{}^\circ }}} \right)\\\\\ \ \ \ \ \ =-\left( {\displaystyle \frac{{1-\displaystyle \frac{1}{{\sqrt{3}}}}}{{1+\displaystyle \frac{1}{{\sqrt{3}}}}}} \right)\\\\\ \ \ \ \ \ =-\left( {\displaystyle \frac{{\sqrt{3}-1}}{{\sqrt{3}+1}}\times \displaystyle \frac{{\sqrt{3}-1}}{{\sqrt{3}-1}}} \right)\\\\\ \ \ \ \ \ =-\left( {\displaystyle \frac{{3-2\sqrt{3}+1}}{2}} \right)\\\\\ \ \ \ \ \ =\sqrt{3}-2\\\\\end{array}$

$ \displaystyle \begin{array}{l} \text{(i)}\ \ \ \ \ \ \cos 345{}^\circ \\\\\ \ \ \ \ \ =\cos (360{}^\circ -15{}^\circ )\\\\\ \ \ \ \ \ =\cos 15{}^\circ \\\\\ \ \ \ \ \ =\cos (45{}^\circ -30{}^\circ )\\\\\ \ \ \ \ \ =\cos 45{}^\circ \cos 30{}^\circ +\sin 45{}^\circ \sin 30{}^\circ \\\\\ \ \ \ \ \ =\displaystyle \frac{{\sqrt{2}}}{2}\times \displaystyle \frac{{\sqrt{3}}}{2}+\displaystyle \frac{{\sqrt{2}}}{2}\times \displaystyle \frac{1}{2}\\\\\ \ \ \ \ \ =\displaystyle \frac{{\sqrt{6}+\sqrt{2}}}{4}\\\\ \text{(j)}\ \ \ \ \ \ \tan 15{}^\circ \\\\\ \ \ \ \ \ =\tan (45{}^\circ -30{}^\circ )\\\\\ \ \ \ \ \ =\displaystyle \frac{{\tan 45{}^\circ -\tan 30{}^\circ }}{{1+\tan 45{}^\circ \tan 30{}^\circ }}\\\\\ \ \ \ \ \ =\displaystyle \frac{{1-\displaystyle \frac{1}{{\sqrt{3}}}}}{{1+\displaystyle \frac{1}{{\sqrt{3}}}}}\\\\\ \ \ \ \ \ =\displaystyle \frac{{\sqrt{3}-1}}{{\sqrt{3}+1}}\times \displaystyle \frac{{\sqrt{3}-1}}{{\sqrt{3}-1}}\\\\\ \ \ \ \ \ =\displaystyle \frac{{3-2\sqrt{3}+1}}{2}\\\\\ \ \ \ \ \ =2-\sqrt{3}\\\\\end{array}$

$ \displaystyle \begin{array}{l} \text{(k)}\ \ \ \ \ \sin (-15{}^\circ )\\\\\ \ \ \ \ \ =-\sin 15{}^\circ \\\\\ \ \ \ \ \ =-\sin (45{}^\circ -30{}^\circ )\\\\\ \ \ \ \ \ =-\left( {\sin 45{}^\circ \cos 30{}^\circ -\cos 45{}^\circ \sin 30{}^\circ } \right)\\\\\ \ \ \ \ \ =-\left( {\displaystyle \frac{{\sqrt{2}}}{2}\times \displaystyle \frac{{\sqrt{3}}}{2}-\displaystyle \frac{{\sqrt{2}}}{2}\times \displaystyle \frac{1}{2}} \right)\\\\\ \ \ \ \ \ =-\left( {\displaystyle \frac{{\sqrt{6}-\sqrt{2}}}{4}} \right)\\\\\ \ \ \ \ \ =\displaystyle \frac{{\sqrt{2}-\sqrt{6}}}{4}\\\\ \text{(l)}\ \ \ \ \ \ \sec (-75{}^\circ )\ \ \\\\\ \ \ \ \ \ =\sec 75{}^\circ \\\\\ \ \ \ \ \ =\displaystyle \frac{1}{{\cos 75{}^\circ }}\\\\\ \ \ \ \ \ =\displaystyle \frac{1}{{\cos (45{}^\circ +30{}^\circ )}}\\\\\ \ \ \ \ \ =\displaystyle \frac{1}{{\cos 45{}^\circ \cos 30{}^\circ -\sin 45{}^\circ \sin 30{}^\circ }}\\\\\ \ \ \ \ \ =\displaystyle \frac{1}{{\displaystyle \frac{{\sqrt{2}}}{2}\times \displaystyle \frac{{\sqrt{3}}}{2}-\displaystyle \frac{{\sqrt{2}}}{2}\times \displaystyle \frac{1}{2}}}\\\\\ \ \ \ \ \ =\displaystyle \frac{4}{{\sqrt{6}-\sqrt{2}}}\times \displaystyle \frac{{\sqrt{6}+\sqrt{2}}}{{\sqrt{6}+\sqrt{2}}}\\\\\ \ \ \ \ \ =\sqrt{6}+\sqrt{2}\end{array}$

2.       Express the following as single trigonometric ratios:

(a) $ \displaystyle \sin 37{}^\circ \cos 41{}^\circ + \cos 37{}^\circ \sin 41{}^\circ$

(b) $ \displaystyle \frac{{1-\tan 15{}^\circ }}{{1+\tan 15{}^\circ }}$

(c) $ \displaystyle \cos 25{}^\circ \cos 15{}^\circ- \sin 25{}^\circ \sin 15{}^\circ$

(d) $ \displaystyle \cos 75{}^\circ \cos 24{}^\circ + \sin 75{}^\circ \sin 24{}^\circ$

(e) $ \displaystyle \sin 126{}^\circ \cos 23{}^\circ- \cos 126{}^\circ \sin 23{}^\circ$

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$ \displaystyle \begin{array}{l}\text{(a)}\ \ \ \ \ \sin 37{}^\circ \cos 41{}^\circ +\cos 37{}^\circ \sin 41{}^\circ \\\\\ \ \ \ \ =\ \sin (37{}^\circ +41{}^\circ )\\\\\ \ \ \ \ =\ \sin 78{}^\circ \\\\\text{(b)}\ \ \ \ \displaystyle \frac{{1-\tan 15{}^\circ }}{{1+\tan 15{}^\circ }}\\\\\ \ \ \ \ =\ \displaystyle \frac{{\tan 45{}^\circ -\tan 15{}^\circ }}{{1+\tan 45{}^\circ \tan 15{}^\circ }}\ \ \ \left[ {\because \tan 45{}^\circ =1} \right]\\\\\ \ \ \ \ =\ \tan (45{}^\circ -15{}^\circ )\\\\\ \ \ \ \ =\ \tan 30{}^\circ \\\\\text{(c)}\ \ \ \ \cos 25{}^\circ \cos 15{}^\circ -\sin 25{}^\circ \sin 15{}^\circ \\\\\ \ \ \ \ =\ \cos (25{}^\circ +15{}^\circ )\\\\\ \ \ \ \ =\ \cos 40{}^\circ \\\\\text{(d)}\ \ \ \ \cos 75{}^\circ \cos 24{}^\circ +\sin 75{}^\circ \sin 24{}^\circ \\\\\ \ \ \ \ =\ \cos (75{}^\circ -24{}^\circ )\\\\\ \ \ \ \ =\ \cos 51{}^\circ \\\\\text{(e)}\ \ \ \ \ \sin 126{}^\circ \cos 23{}^\circ -\cos 126{}^\circ \sin 23{}^\circ \\\\\ \ \ \ \ =\ \sin (126{}^\circ -23{}^\circ )\\\\\ \ \ \ \ =\ \sin 103{}^\circ \end{array}$

3.       (a) Express $ \displaystyle \cos 3x$ in terms of $ \displaystyle \cos x.$

(b) Express $ \displaystyle \sin 3x$ in terms of $ \displaystyle \sin x.$

(c) Express $ \displaystyle \tan 3x$ in terms of $ \displaystyle \tan x.$

(d) Express $ \displaystyle \cot 2x$ in terms of $ \displaystyle \cot x.$

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$ \displaystyle \begin{array}{l}\text{(a)}\ \ \ \ \ \cos 3x\\\\\ \ \ \ \ =\ \ \cos (2x+x)\\\\\ \ \ \ \ =\ \ \cos 2x\cos x-\sin 2x\sin x\\\\\ \ \ \ \ =\ \ (2{{\cos }^{2}}x-1)\cos x-(2\sin x\cos x)\sin x\\\\\ \ \ \ \ =\ \ 2{{\cos }^{3}}x-\cos x-2{{\sin }^{2}}x\cos x\\\\\ \ \ \ \ =\ \ 2{{\cos }^{3}}x-\cos x-2(1-{{\cos }^{2}}x)\cos x\\\\\ \ \ \ \ =\ \ 2{{\cos }^{3}}x-\cos x-2\cos x+2{{\cos }^{3}}x\\\\\ \ \ \ \ =\ \ 4{{\cos }^{3}}x-3\cos x\\\\\text{(b)}\ \ \ \ \ \sin 3x\\\\\ \ \ \ \ =\ \ \sin (2x+x)\\\\\ \ \ \ \ =\ \ \sin 2x\cos x+\cos 2x\sin x\\\\\ \ \ \ \ =\ \ (2\sin x\cos x)\cos x+(1-2{{\sin }^{2}}x)\sin x\\\\\ \ \ \ \ =\ \ 2\sin x{{\cos }^{2}}x+\sin x-2{{\sin }^{3}}x\\\\\ \ \ \ \ =\ \ 2\sin x(1-{{\sin }^{2}}x)+\sin x-2{{\sin }^{3}}x\\\\\ \ \ \ \ =\ \ 2\sin x-2{{\sin }^{3}}x+\sin x-2{{\sin }^{3}}x\\\\\ \ \ \ \ =\ \ 3\sin x-4{{\sin }^{3}}x\\\\\text{(c)}\ \ \ \ \ \tan 3x\\\\\ \ \ \ \ =\ \ \tan (2x+x)\\\\\ \ \ \ \ =\ \ \displaystyle \frac{{\tan 2x+\tan x}}{{1-\tan 2x\tan x}}\\\\\ \ \ \ \ =\ \ \displaystyle \frac{{\displaystyle \frac{{2\tan x}}{{1-{{{\tan }}^{2}}x}}+\tan x}}{{1-\displaystyle \frac{{2\tan x}}{{1-{{{\tan }}^{2}}x}}\times \tan x}}\\\\\ \ \ \ \ =\ \ \displaystyle \frac{{\displaystyle \frac{{2\tan x+\tan x-{{{\tan }}^{3}}x}}{{1-{{{\tan }}^{2}}x}}}}{{\displaystyle \frac{{1-{{{\tan }}^{2}}x-2{{{\tan }}^{2}}x}}{{1-{{{\tan }}^{2}}x}}}}\\\\\ \ \ \ \ =\ \ \displaystyle \frac{{3\tan x-{{{\tan }}^{3}}x}}{{1-{{{\tan }}^{2}}x}}\times \displaystyle \frac{{1-{{{\tan }}^{2}}x}}{{1-3{{{\tan }}^{2}}x}}\\\\\ \ \ \ \ =\ \ \displaystyle \frac{{3\tan x-{{{\tan }}^{3}}x}}{{1-3{{{\tan }}^{2}}x}}\\\\\text{(d)}\ \ \ \ \ \cot 2x\\\\\ \ \ \ \ =\ \ \displaystyle \frac{1}{{\tan 2x}}\\\\\ \ \ \ \ =\ \ \displaystyle \frac{1}{{\displaystyle \frac{{2\tan x}}{{1-{{{\tan }}^{2}}x}}}}\\\\\ \ \ \ \ =\ \ \displaystyle \frac{{1-{{{\tan }}^{2}}x}}{{2\tan x}}\\\\\ \ \ \ \ =\ \ \displaystyle \frac{{1-\displaystyle \frac{1}{{{{{\cot }}^{2}}x}}}}{{\displaystyle \frac{2}{{\cot x}}}}\\\\\ \ \ \ \ =\ \displaystyle \frac{{{{{\cot }}^{2}}x-1}}{{{{{\cot }}^{2}}x}}\times \displaystyle \frac{{\cot x}}{2}\\\\\ \ \ \ \ =\ \displaystyle \frac{{{{{\cot }}^{2}}x-1}}{{2\cot x}}\end{array}$

4.       If $ \displaystyle \tan (x - y) = \frac{{25}}{{24}}$ and $ \displaystyle \tan x = \frac{3}{4},$ find the exact value of $ \displaystyle \tan y.$

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \tan (x-y)= \displaystyle \frac{{25}}{{24}},\tan x= \displaystyle \frac{3}{4}\\\\\therefore \ \ \ \displaystyle \frac{{\tan x-\tan y}}{{1+\tan x\tan y}}= \displaystyle \frac{{25}}{{24}}\\\\\therefore \ \ \ \displaystyle \frac{{ \displaystyle \frac{3}{4}-\tan y}}{{1+ \displaystyle \frac{3}{4}\tan y}}= \displaystyle \frac{{25}}{{24}}\\\\\therefore \ \ \ 18-24\tan y=25+ \displaystyle \frac{{75}}{4}\tan y\\\\\therefore \ \ \ \displaystyle \frac{{75}}{4}\tan y+24\tan y=7\\\\\therefore \ \ \ \displaystyle \frac{{171}}{4}\tan y=7\\\\\therefore \ \ \ \tan y= \displaystyle \frac{{28}}{{171}}\end{array}$

5.       Without the use of table evaluate $ \displaystyle \tan ( \alpha + \beta + \gamma ),$ given that $ \displaystyle \tan \alpha= \frac{1}{2}, \tan \beta= \frac{1}{3}, \tan \gamma= \frac{1}{4}.$

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \tan \alpha = \displaystyle \frac{1}{2},\ \ \tan \beta = \displaystyle \frac{1}{3},\ \ \tan \gamma = \displaystyle \frac{1}{4}\\\\\therefore \ \ \ \tan (\alpha +\beta )= \displaystyle \frac{{\tan \alpha +\tan \beta }}{{1-\tan \alpha \tan \beta }}\\\\\therefore \ \ \ \tan (\alpha +\beta )= \displaystyle \frac{{ \displaystyle \frac{1}{2}+ \displaystyle \frac{1}{3}}}{{1- \displaystyle \frac{1}{2}\times \displaystyle \frac{1}{3}}}\\\\\therefore \ \ \ \tan (\alpha +\beta )= \displaystyle \frac{{ \displaystyle \frac{{3+2}}{6}}}{{1- \displaystyle \frac{1}{6}}}\\\\\therefore \ \ \ \tan (\alpha +\beta )= \displaystyle \frac{{ \displaystyle \frac{5}{6}}}{{ \displaystyle \frac{5}{6}}}=1\\\\\therefore \ \ \ \tan (\alpha +\beta +\gamma )=\tan \left[ {(\alpha +\beta )+\gamma } \right]\\\\\therefore \ \ \ \tan (\alpha +\beta +\gamma )= \displaystyle \frac{{\tan (\alpha +\beta )+\tan \gamma }}{{1-\tan (\alpha +\beta )\tan \gamma }}\\\\\therefore \ \ \ \tan (\alpha +\beta +\gamma )= \displaystyle \frac{{1+ \displaystyle \frac{1}{4}}}{{1- \displaystyle \frac{1}{4}}}= \displaystyle \frac{5}{3}\end{array}$

6.       Given that $ \displaystyle \tan \alpha= \frac{1}{3}, \tan \beta= \frac{1}{4}, \tan \gamma= \frac{1}{6}.$ Without the use of table evaluate $ \displaystyle \tan ( \alpha + \beta - \gamma ).$

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \tan \alpha = \displaystyle \frac{1}{3},\ \ \tan \beta = \displaystyle \frac{1}{4},\ \ \tan \gamma = \displaystyle \frac{1}{6}\\\\\therefore \ \ \ \tan (\alpha +\beta )= \displaystyle \frac{{\tan \alpha +\tan \beta }}{{1-\tan \alpha \tan \beta }}\\\\\therefore \ \ \ \tan (\alpha +\beta )= \displaystyle \frac{{ \displaystyle \frac{1}{3}+ \displaystyle \frac{1}{4}}}{{1- \displaystyle \frac{1}{3}\times \displaystyle \frac{1}{4}}}\\\\\therefore \ \ \ \tan (\alpha +\beta )= \displaystyle \frac{{ \displaystyle \frac{{4+3}}{{12}}}}{{1- \displaystyle \frac{1}{{12}}}}\\\\\therefore \ \ \ \tan (\alpha +\beta )= \displaystyle \frac{{ \displaystyle \frac{7}{{12}}}}{{ \displaystyle \frac{{11}}{{12}}}}= \displaystyle \frac{7}{{11}}\\\\\therefore \ \ \ \tan (\alpha +\beta +\gamma )=\tan \left[ {(\alpha +\beta )-\gamma } \right]\\\\\therefore \ \ \ \tan (\alpha +\beta -\gamma )= \displaystyle \frac{{\tan (\alpha +\beta )-\tan \gamma }}{{1+\tan (\alpha +\beta )\tan \gamma }}\\\\\therefore \ \ \ \tan (\alpha +\beta -\gamma )= \displaystyle \frac{{ \displaystyle \frac{7}{{11}}- \displaystyle \frac{1}{6}}}{{1+ \displaystyle \frac{7}{{11}}\times \displaystyle \frac{1}{6}}}\\\\\therefore \ \ \ \tan (\alpha +\beta -\gamma )= \displaystyle \frac{{ \displaystyle \frac{{31}}{{66}}}}{{ \displaystyle \frac{{73}}{{66}}}}= \displaystyle \frac{{31}}{{73}}\end{array}$

7.       Given that $ \displaystyle \cos 2A =\frac{{119}}{{169}}$ and that $ \displaystyle A$ is acute, find, without using tables, the value of $ \displaystyle \tan 2A$ and $ \displaystyle \cos A.$

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \cos 2A= \displaystyle \frac{{119}}{{169}}\\\\\therefore \ \ \ \ 2{{\cos }^{2}}A-1= \displaystyle \frac{{119}}{{169}}\\\\\therefore \ \ \ \ 2{{\cos }^{2}}A= \displaystyle \frac{{288}}{{169}}\\\\\therefore \ \ \ \ {{\cos }^{2}}A= \displaystyle \frac{{144}}{{169}}\\\\\therefore \ \ \ \ \cos A= \displaystyle \frac{{12}}{{13}}\ \ \ \left[ {\because A\ \text{is acute}} \right]\\\\\therefore \ \ \ \ \sin A=\sqrt{{1-{{{\cos }}^{2}}A}}\ \ \ \left[ {\because {{{\sin }}^{2}}A+{{{\cos }}^{2}}A=1} \right]\\\\\therefore \ \ \ \ \sin A=\sqrt{{1- \displaystyle \frac{{144}}{{169}}}}\ = \displaystyle \frac{5}{{13}}\\\\\therefore \ \ \ \ \sin 2A=2\sin A\cos A=2\ \left( { \displaystyle \frac{5}{{13}}} \right)\left( { \displaystyle \frac{{12}}{{13}}} \right)= \displaystyle \frac{{120}}{{169}}\\\\\therefore \ \ \ \ \tan 2A= \displaystyle \frac{{\sin 2A}}{{\cos 2A}}= \displaystyle \frac{{ \displaystyle \frac{{120}}{{169}}}}{{ \displaystyle \frac{{119}}{{169}}}}= \displaystyle \frac{{120}}{{119}}\end{array}$

8.       Given that $ \displaystyle \tan α = p$ and $ \displaystyle \tan (α -β ) = q,$ express $ \displaystyle \tan β$ in terms of $ \displaystyle p$ and $ \displaystyle q.$ Calculate the values of $ \displaystyle \tan (α + β )$ when $ \displaystyle p = 1$ and $ \displaystyle q = 0.5.$

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \tan \alpha =p\\\\\ \ \ \ \ \tan (\alpha -\beta )=q\\\\\therefore \ \ \ \displaystyle \frac{{\tan \alpha -\tan \beta }}{{1+\tan \alpha \tan \beta }}=q\\\\\therefore \ \ \ \displaystyle \frac{{p-\tan \beta }}{{1+p\tan \beta }}=q\\\\\therefore \ \ \ p-\tan \beta =q+pq\tan \beta \\\\\therefore \ \ \ \tan \beta +pq\tan \beta =p-q\\\\\therefore \ \ \ \tan \beta (1+pq)=p-q\\\\\therefore \ \ \ \tan \beta = \displaystyle \frac{{p-q}}{{1+pq}}\\\\\ \ \ \ \ \text{When}\ p=1,q=0.5,\\\\\ \ \ \ \ \tan \alpha =p=1\\\\\ \ \ \ \ \tan \beta = \displaystyle \frac{{p-q}}{{1+pq}}= \displaystyle \frac{{1-0.5}}{{1+0.5}}= \displaystyle \frac{1}{3}\\\\\ \ \ \ \ \tan (\alpha +\beta )= \displaystyle \frac{{\tan \alpha +\tan \beta }}{{1-\tan \alpha \tan \beta }}\\\\\therefore \ \ \ \tan (\alpha +\beta )= \displaystyle \frac{{1+ \displaystyle \frac{1}{3}}}{{1- \displaystyle \frac{1}{3}}}= \displaystyle \frac{{ \displaystyle \frac{4}{3}}}{{ \displaystyle \frac{2}{3}}}=2\end{array}$

9.       If $ \displaystyle \sin \theta = a,$ where $ \displaystyle \sin \theta$ is an acute angle express the following in terms of $ \displaystyle a:$

(a) $ \displaystyle \tan^2 \theta \ \ \ \ \ \ $ (b) $ \displaystyle \cos 2 \theta \ \ \ \ \ \ $ (c) $ \displaystyle \sin 4 \theta \ \ \ \ \ \ $ (d) $ \displaystyle\cos^2 \frac{1}{2}\theta.$

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \sin \theta =a,\ 0{}^\circ <\theta <90{}^\circ \\\\\therefore \ \ \ \cos \theta =\sqrt{{1-{{a}^{2}}}}\\\\\therefore \ \ \ \tan \theta = \displaystyle \frac{{\sin \theta }}{{\cos \theta }}= \displaystyle \frac{a}{{\sqrt{{1-{{a}^{2}}}}}}\\\\\therefore \ \ \ {{\tan }^{2}}\theta = \displaystyle \frac{{a}^{2}}{{1-{{a}^{2}}}}\\\\\therefore \ \ \ \cos 2\theta =1-2{{\sin }^{2}}\theta \\\\\therefore \ \ \ \cos 2\theta =1-2{{a}^{2}}\\\\\ \ \ \ \sin 4\theta =\sin 2(2\theta )\\\\\therefore \ \ \sin 4\theta =2\sin 2\theta \cos 2\theta \\\\\therefore \ \ \sin 4\theta =2(2\sin \theta \cos \theta )\cos 2\theta \\\\\therefore \ \ \sin 4\theta =4\sin \theta \cos \theta \cos 2\theta \\\\\therefore \ \ \sin 4\theta =4a\sqrt{{1-{{a}^{2}}}}(1-2{{a}^{2}})\\\\\ \ \ \ \cos \displaystyle \frac{1}{2}\theta =\sqrt{{ \displaystyle \frac{{1+\cos \theta }}{2}}}\\\\\therefore \ \ {{\cos }^{2}} \displaystyle \frac{1}{2}\theta = \displaystyle \frac{{1+\cos \theta }}{2}= \displaystyle \frac{{1+\sqrt{{1-{{a}^{2}}}}}}{2}\ \end{array}$

10.    Given that $ \displaystyle \alpha$ is an acute angle and $ \displaystyle \cos \alpha = x,$ find, in terms of $ \displaystyle x$ the value of

(a) $ \displaystyle \tan^2 \alpha \ \ \ \ \ \ $ (b) $ \displaystyle \sin 2\alpha \ \ \ \ \ \ $ (c) $ \displaystyle \cos 4\alpha \ \ \ \ \ \ $ (d) $ \displaystyle \sin \frac{1}{2}\alpha.$

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \cos \alpha =x,\ 0{}^\circ <\alpha <90{}^\circ \\\\\therefore \ \ \ \sin \alpha =\sqrt{{1-{{x}^{2}}}}\\\\\therefore \ \ \ \tan \alpha = \displaystyle \frac{{\sin \alpha }}{{\cos \alpha }}= \displaystyle \frac{{\sqrt{{1-{{x}^{2}}}}}}{x}\\\\\therefore \ \ \ {{\tan }^{2}}\alpha = \displaystyle \frac{{1-{{x}^{2}}}}{{{{x}^{2}}}}\\\\\therefore \ \ \ \sin 2\alpha =2\sin \alpha \cos \alpha \\\\\therefore \ \ \ \sin 2\alpha =2x\sqrt{{1-{{x}^{2}}}}\\\\\ \ \ \ \cos 4\alpha =\cos 2(2\alpha )\\\\\therefore \ \ \cos 4\alpha =1-2{{\sin }^{2}}2\alpha \\\\\therefore \ \ \cos 4\alpha =1-2{{\left( {2x\sqrt{{1-{{x}^{2}}}}} \right)}^{2}}\\\\\therefore \ \ \cos 4\alpha =1-2\left[ {4{{x}^{2}}(1-{{x}^{2}})} \right]\\\\\therefore \ \ \cos 4\alpha =1-8{{x}^{2}}+8{{x}^{4}}\\\\\therefore \ \ \sin \displaystyle \frac{1}{2}\alpha =\sqrt{{ \displaystyle \frac{{1-\cos \alpha }}{2}}}\\\\\therefore \ \ \sin \displaystyle \frac{1}{2}\alpha =\sqrt{{ \displaystyle \frac{{1-x}}{2}}}\ \ \ \ \ \left[ {\because 0{}^\circ <\alpha <90{}^\circ } \right]\end{array}$

11.    If $ \displaystyle \sin \theta = a,$ where $ \displaystyle \theta$ is an acute angle express the following in terms of $ \displaystyle a:$

(a) $ \displaystyle \cot 2\theta \ \ \ \ \ \ $ (b) $ \displaystyle \sec 2\theta \ \ \ \ \ \ $ (c) $ \displaystyle \operatorname{cosec} 2\theta.$

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \sin \theta =a,\ 0{}^\circ <\theta <90{}^\circ \\\\\therefore \ \ \ \cos \theta =\sqrt{{1-{{a}^{2}}}}\\\\\ \ \ \ \ \sin 2\theta =2\sin \theta \cos \theta \\\\\therefore \ \ \ \sin 2\theta =2a\sqrt{{1-{{a}^{2}}}}\\\\\ \ \ \ \ \cos 2\theta =1-2{{\sin }^{2}}\theta \\\\\therefore \ \ \ \cos 2\theta =1-2{{a}^{2}}\\\\\therefore \ \ \ \cot 2\theta = \displaystyle \frac{{\cos 2\theta }}{{\sin 2\theta }}= \displaystyle \frac{{1-2{{a}^{2}}}}{{2a\sqrt{{1-{{a}^{2}}}}}}\\\\\therefore \ \ \ \sec 2\theta = \displaystyle \frac{1}{{\cos 2\theta }}= \displaystyle \frac{1}{{1-2{{a}^{2}}}}\\\\\therefore \ \ \ \operatorname{cosec}2\theta = \displaystyle \frac{1}{{\sin 2\theta }}= \displaystyle \frac{1}{{2a\sqrt{{1-{{a}^{2}}}}}}\end{array}$

12.    Given that $ \displaystyle \sin \alpha = -\frac{4}{5}, \cos \beta = - \frac{12}{13}$ and that $ \displaystyle \alpha$ and $ \displaystyle \beta$ are in the same quadrant, find each of the following without the use of tables.

(a) $ \displaystyle \sin 2\alpha \ \ \ \ \ \ $ (b) $ \displaystyle \cos 2\alpha \ \ \ \ \ \ $ (c) $ \displaystyle \cos\frac {\beta}{2}\ \ \ \ \ \ $ (c) $ \displaystyle \tan 2 \beta.$

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$ \displaystyle \begin{array}{l}\ \ \ \ \sin \alpha =-\displaystyle \frac{4}{5},\ \ \ \cos \beta =-\displaystyle \frac{{12}}{{13}}\\\ \ \ \ \text{and }\alpha \ \text{and}\ \beta \ \text{are in the same quadrant}\text{.}\\\\\ \ \ \ \text{Since both}\sin \alpha \ \text{ and}\cos \beta \ \text{are negative,}\\\ \ \ \ \alpha \ \text{and}\ \beta \text{ will be in }{{\text{3}}^{{\text{rd}}}}\text{ quadrant}\text{.}\end{array}$

$ \displaystyle \therefore \ \ \ \sin \alpha =-\frac{4}{5},\ \cos \alpha =-\frac{3}{5}$

$ \displaystyle \begin{array}{l}\therefore \ \ \ \sin \beta =-\displaystyle \frac{5}{{13}},\ \\\\\ \ \ \ \ \cos \beta =-\displaystyle \frac{{12}}{{13}},\\\\\ \ \ \ \ \tan \beta =\displaystyle \frac{5}{{12}}\\\\\\\ \ \ \ \ \sin 2\alpha =2\sin \alpha \cos \alpha \\\\\therefore \ \ \ \sin 2\alpha =2\left( {-\displaystyle \frac{4}{5}} \right)\left( {-\displaystyle \frac{3}{5}} \right)\\\\\therefore \ \ \ \sin 2\alpha =\displaystyle \frac{{24}}{{25}}\\\\\ \ \ \ \ \cos 2\alpha =2{{\cos }^{2}}\alpha -1\\\\\therefore \ \ \ \cos 2\alpha =2{{\left( {-\displaystyle \frac{3}{5}} \right)}^{2}}-1\\\\\therefore \ \ \ \cos 2\alpha =-\displaystyle \frac{7}{{25}}\\\\\ \ \ \ \cos \displaystyle \frac{\beta }{2}=\pm \sqrt{{\displaystyle \frac{{1+\cos \beta }}{2}}}\\\\\therefore \ \ \cos \displaystyle \frac{\beta }{2}=\pm \sqrt{{\displaystyle \frac{{1-\displaystyle \frac{{12}}{{13}}}}{2}}}\\\\\therefore \ \ \cos \displaystyle \frac{\beta }{2}=\pm \sqrt{{\displaystyle \frac{1}{{26}}}}=\pm \displaystyle \frac{{\sqrt{{26}}}}{{26}}\\\\\ \ \ \ \text{Since}\ 180{}^\circ <\beta <270{}^\circ ,\\\\\ \ \ \ 90{}^\circ <\displaystyle \frac{\beta }{2}<135{}^\circ ,\\\\\therefore \ \ \cos \displaystyle \frac{\beta }{2}=-\displaystyle \frac{{\sqrt{{26}}}}{{26}}\\\\\ \ \ \ \tan 2\beta =\displaystyle \frac{{2\tan \beta }}{{1-{{{\tan }}^{2}}\beta }}\\\\\therefore \ \ \tan 2\beta =\displaystyle \frac{{2\left( {\displaystyle \frac{5}{{12}}} \right)}}{{1-{{{\left( {\displaystyle \frac{5}{{12}}} \right)}}^{2}}}}\\\\\therefore \ \ \tan 2\beta =\displaystyle \frac{{\displaystyle \frac{5}{6}}}{{\displaystyle \frac{{119}}{{144}}}}=\displaystyle \frac{5}{6}\times \displaystyle \frac{{144}}{{119}}=\displaystyle \frac{{120}}{{119}}\end{array}$

13.    Given that $ \displaystyle \sin \alpha = \frac{15}{17}, \cos \beta = \frac{3}{5}$ and that $ \displaystyle \alpha$ and $ \displaystyle \beta$ are in the same quadrant, find without using tables, the value of

(a) $ \displaystyle \sin 2\alpha \ \ \ \ \ \ $ (b)$ \displaystyle \cos\frac {\alpha}{2}\ \ \ \ \ \ $ (c) $ \displaystyle \cos 2 \beta.$

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$ \displaystyle \begin{array}{l}\ \ \ \ \sin \alpha = \displaystyle \frac{{15}}{{17}},\ \ \ \cos \beta =- \displaystyle \frac{3}{5}\\\ \ \ \ \text{and }\alpha \ \text{and}\ \beta \ \text{are in the same quadrant}\text{.}\\\\\ \ \ \ \text{Since}\sin \alpha \ \text{is positive and}\cos \beta \ \text{is negative,}\\\ \ \ \ \alpha \ \text{and}\ \beta \ \text{are will be in }{{\text{2}}^{{\text{nd}}}}\text{ quadrant}\text{.}\end{array}$

$ \displaystyle \begin{array}{l}\therefore \ \ \sin \alpha =\displaystyle \frac{{15}}{{17}},\ \cos \alpha =-\displaystyle \frac{8}{{17}}\\\\\therefore \ \ \sin 2\alpha =2\sin \alpha \ \cos \alpha \\\\\ \ \ \ \ \ \ \ \ \ \ \ \ =2\ \left( {\displaystyle \frac{{15}}{{17}}} \right)\left( {-\displaystyle \frac{8}{{17}}} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ =-\displaystyle \frac{{240}}{{289}}\\\\\therefore \ \ \cos \displaystyle \frac{\alpha }{2}=\pm \sqrt{{\displaystyle \frac{{1+\cos \alpha }}{2}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ =\pm \sqrt{{\displaystyle \frac{{1-\displaystyle \frac{8}{{17}}}}{2}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ =\pm \displaystyle \frac{3}{{\sqrt{{34}}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ =\pm \displaystyle \frac{{3\sqrt{{34}}}}{{34}}\\\\\ \ \ \text{Since}\ {{90}^{{}^\circ }}<\alpha <{{180}^{{}^\circ }},{{45}^{{}^\circ }}<\displaystyle \frac{\alpha }{2}<{{90}^{{}^\circ }},\\\\\therefore \ \ \cos \displaystyle \frac{\alpha }{2}=\displaystyle \frac{{3\sqrt{{34}}}}{{34}}\\\\\ \ \ \ \cos \beta =-\displaystyle \frac{3}{5}\ \ \ \ \ \left[ {\text{given}} \right]\\\\\ \ \ \ \cos 2\beta =2{{\cos }^{2}}\beta -1\\\\\therefore \ \ \cos 2\beta =2{{\left( {-\displaystyle \frac{3}{5}} \right)}^{2}}-1\\\\\therefore \ \ \cos 2\beta =-\displaystyle \frac{7}{{25}}\end{array}$

14.    Given that $ \displaystyle \tan \theta =\frac{3}{4}, \cos \phi = - \frac{3}{5}$ and that $ \displaystyle \theta$ and $ \displaystyle \phi$ are in the same quadrant, evaluate, $ \displaystyle \frac{{\tan \phi}}{{\cos \theta}}.$

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \tan \theta =\displaystyle \frac{3}{4},\cos \phi =\displaystyle -\frac{7}{{25}}\ \text{and}\ \\\ \ \ \ \ \ \theta \ \text{and}\ \phi \ \text{are in the same quadrant}\text{.}\\\\\ \ \ \ \ \ \text{Since }\tan \theta \ \text{is positive and }\cos \phi \ \text{is negative,}\\\ \ \ \ \ \ \theta \ \text{and }\phi \ \text{will be in the }{{\text{3}}^{{\text{rd}}}}\text{ quadrant}\text{.}\ \end{array}$

$ \displaystyle \therefore \ \ \ \ \sin \theta =-\frac{3}{5}\ \ \text{and}\ \cos \theta =-\frac{4}{5}$

$ \displaystyle \begin{array}{l}\therefore \ \ \ \ \sin \phi =-\displaystyle \frac{{24}}{{25}}\ \ \text{and}\ \tan \phi =\displaystyle \frac{7}{{24}}\\\\\therefore \ \ \ \ \displaystyle \frac{{\tan \phi }}{{\cos \theta }}=\displaystyle \frac{{\displaystyle \frac{7}{{24}}}}{{-\displaystyle \frac{4}{5}}}=\displaystyle \frac{7}{{24}}\left( {-\displaystyle \frac{4}{5}} \right)=-\displaystyle \frac{7}{{30}}\end{array}$

15.    Given that $ \displaystyle \sin \alpha = \frac{3}{5}$ and $ \displaystyle \cos \beta = - \frac{12}{13},$ where $ \displaystyle \alpha$ and $ \displaystyle \beta$ are in the same quadrant. Find the values of $ \displaystyle \sin (\alpha + \beta)$ and $ \displaystyle \cos (\alpha + \beta).$

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \sin \alpha =\displaystyle \frac{3}{5},\cos \beta =-\displaystyle \frac{{12}}{{13}}\ \text{and}\ \\\ \ \ \ \ \ \alpha \ \text{and}\ \beta \ \text{are in the same quadrant}\text{.}\\\\\ \ \ \ \ \ \text{Since }\sin \alpha \ \text{is positive and }\cos \phi \ \text{is negative,}\\\ \ \ \ \ \ \alpha \ \text{and }\beta \ \text{will be in the }{{\text{2}}^{{\text{nd}}}}\text{ quadrant}\text{.}\ \end{array}$

$ \displaystyle \therefore \ \ \ \ \ \sin \alpha =\frac{3}{5}\ \text{and}\ \cos \alpha =-\frac{4}{5}$

$ \displaystyle \begin{array}{l}\therefore \ \ \ \ \ \sin \beta =\displaystyle \frac{5}{{13}}\ \ \text{and}\ \cos \beta =-\displaystyle \frac{{12}}{{13}}\\\\\ \ \ \ \ \ \ \sin \left( {\alpha +\beta } \right)=\sin \alpha \cos \beta +\cos \alpha \sin \beta \ \\\ \\\therefore \ \ \ \ \ \sin \left( {\alpha +\beta } \right)=\displaystyle \frac{3}{5}\left( {-\displaystyle \frac{{12}}{{13}}} \right)+\left( {-\displaystyle \frac{4}{5}} \right)\left( {\displaystyle \frac{5}{{13}}} \right)\\\\\therefore \ \ \ \ \ \sin \left( {\alpha +\beta } \right)=-\displaystyle \frac{{56}}{{65}}\\\\\ \ \ \ \ \ \ \cos \left( {\alpha +\beta } \right)=\cos \alpha \cos \beta -\sin \alpha \sin \beta \\\ \ \\\therefore \ \ \ \ \ \cos \left( {\alpha +\beta } \right)=\left( {-\displaystyle \frac{4}{5}} \right)\left( {-\displaystyle \frac{{12}}{{13}}} \right)-\left( {\displaystyle \frac{3}{5}} \right)\left( {\displaystyle \frac{5}{{13}}} \right)\\\\\therefore \ \ \ \ \ \cos \left( {\alpha +\beta } \right)=\displaystyle \frac{{33}}{{65}}\end{array}$

16.    Given that $ \displaystyle \tan x =\frac{4}{3}, \sin y = -\frac{5}{3}$ and that $ \displaystyle x$ and $ \displaystyle y$ are in the same quadrant, find without using tables, the value of $ \displaystyle \frac{{\cos \ (x-y)}}{{\sin \ (y-x)}}.$

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$ \displaystyle \begin{array}{l}\ \ \ \ \tan x=\displaystyle \frac{4}{3},\ \sin y=-\displaystyle \frac{3}{5}\ \ \text{ and }x\text{ and }y\text{ are }\\\ \ \ \ \ \text{in the same quadrant}\text{.}\\\\\ \ \ \ \ \text{Since }\tan x\text{ is positive and }\sin y\text{ is negative, }\\\ \ \ \ \ x\text{ and }y\text{ will be in the }{{\text{3}}^{{\text{rd}}}}\text{ quadrant}\text{. }\end{array}$

$ \displaystyle \therefore \ \ \ \sin x=-\frac{4}{5}\text{, }\cos x=-\frac{3}{5}$

$ \displaystyle \begin{array}{l}\therefore \ \ \ \ \ \ \ \sin y=-\displaystyle \frac{3}{5}\text{, }\cos y=-\displaystyle \frac{4}{5}\\\\\therefore \ \ \ \ \ \ \ \cos \left( {x-y} \right)\\\\\ \ \ \ =\ \ \ \cos x\cos y+\sin x\sin y\\\\\ \ \ \ =\ \ \ \left( {-\displaystyle \frac{3}{5}} \right)\left( {-\displaystyle \frac{4}{5}} \right)+\left( {-\displaystyle \frac{4}{5}} \right)\left( {-\displaystyle \frac{3}{5}} \right)\\\\\ \ \ \ =\ \ \ \displaystyle \frac{{24}}{{25}}\ \\\\\therefore \ \ \ \ \ \ \ \sin \left( {y-x} \right)\\\\\ \ \ \ =\ \ \ \sin y\cos x-\cos y\sin x\\\\\ \ \ \ =\ \ \ \left( {-\displaystyle \frac{3}{5}} \right)\left( {-\displaystyle \frac{3}{5}} \right)-\left( {-\displaystyle \frac{4}{5}} \right)\left( {-\displaystyle \frac{4}{5}} \right)\\\ \ \ \ \\\therefore \ \ \ \displaystyle \frac{{\cos \left( {x-y} \right)}}{{\sin \left( {y-x} \right)}}=\displaystyle \frac{{\displaystyle \frac{{24}}{{25}}}}{{-\displaystyle \frac{7}{{25}}}}=\ -\displaystyle \frac{{24}}{7}\ \text{ }\end{array}$

17.    Given that $ \displaystyle \sin \phi =\frac{12}{13}$ and $ \displaystyle \cos \theta =\frac{4}{5},$ where $ \displaystyle \phi$ and $ \displaystyle \theta$ are in the same quadrant. Without using tables, find the values of $ \displaystyle \tan (\phi + \theta)$ and $ \displaystyle \tan (\phi - \theta).$

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \sin \phi =\displaystyle \frac{{12}}{{13}},\ \cos \theta =\displaystyle \frac{4}{5}\ \ \text{ and }\phi \text{ and }\theta \text{ are }\\\ \ \ \ \ \text{in the same quadrant}\text{.}\\\\\ \ \ \ \ \text{Since both }\sin \phi \text{ and}\cos \theta \text{ are positive, }\\\ \ \ \ \ \phi \text{ and }\theta \text{ will be in the }{{\text{1}}^{{\text{st}}}}\text{ quadrant}\text{.}\end{array}$

$ \displaystyle \therefore \ \ \ \ \ \ \ \cos \phi =\frac{5}{{13}}\text{, }\tan \phi =\frac{{12}}{5}$

$ \displaystyle \begin{array}{l}\therefore \ \ \ \ \ \ \ \sin \theta =\displaystyle \frac{3}{5}\text{, }\tan \theta =\displaystyle \frac{3}{4}\\\\\therefore \ \ \ \ \ \ \ \tan \left( {\phi +\theta } \right)\\\\\ \ \ \ =\ \ \ \displaystyle \frac{{\tan \phi +\tan \theta }}{{1-\tan \phi \tan \theta }}\\\\\ \ \ \ =\ \ \ \displaystyle \frac{{\displaystyle \frac{{12}}{5}+\displaystyle \frac{3}{4}}}{{1-\left( {\displaystyle \frac{{12}}{5}\times \displaystyle \frac{3}{4}} \right)}}\\\\\ \ \ \ =\ \ \ \displaystyle \frac{{\displaystyle \frac{{63}}{{20}}}}{{-\displaystyle \frac{{16}}{{20}}}}\\\\\ \ \ \ =\ \ -\ \displaystyle \frac{{63}}{{16}}\\\\\ \ \ \therefore \ \ \ \ \ \ \ \tan \left( {\phi -\theta } \right)\\\\\ \ \ \ =\ \ \ \displaystyle \frac{{\tan \phi -\tan \theta }}{{1+\tan \phi \tan \theta }}\\\\\ \ \ \ =\ \ \ \displaystyle \frac{{\displaystyle \frac{{12}}{5}-\displaystyle \frac{3}{4}}}{{1+\left( {\displaystyle \frac{{12}}{5}\times \displaystyle \frac{3}{4}} \right)}}\\\\\ \ \ \ =\ \ \ \displaystyle \frac{{\displaystyle \frac{{33}}{{20}}}}{{\displaystyle \frac{{56}}{{20}}}}\\\\\ \ \ \ =\ \ \ \displaystyle \frac{{33}}{{56}}\end{array}$

18.    Given that $ \displaystyle \tan \theta =\frac{3}{4} , 180{}^\circ < \theta < 270{}^\circ$ and $ \displaystyle \sin \phi = -\frac{4}{5} , 270{}^\circ <\phi< 360{}^\circ$ find the values of $ \displaystyle \sin \frac{\phi}{2}$ and $ \displaystyle \sin (\theta + \phi).$

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \tan \theta =\displaystyle \frac{3}{4}\ \text{and 180}{}^\circ <\theta <\text{270}{}^\circ .\\\\\therefore \ \ \ \theta \ \text{is}\ \text{in the }{{\text{3}}^{{\text{rd}}}}\text{ quadrant}\text{.}\end{array}$

$ \displaystyle \therefore \ \ \ \sin \theta =-\frac{3}{5}\ \ \text{and}\ \cos \theta =-\frac{4}{5}$

$ \displaystyle \begin{array}{l}\ \ \ \ \ \sin \phi =-\displaystyle \frac{4}{5}\ \text{and 270}{}^\circ <\theta <\text{360}{}^\circ .\\\\\therefore \ \ \ \phi \ \text{is}\ \text{in the }{{\text{4}}^{{\text{th}}}}\text{ quadrant}\text{.}\end{array}$

$ \displaystyle \begin{array}{l}\therefore \ \ \ \sin \phi =-\displaystyle \frac{4}{5}\ \ \text{and}\ \cos \phi =\displaystyle \frac{3}{5}\\\\\ \ \ \ \ \sin \displaystyle \frac{\phi }{2}=\pm \sqrt{{\displaystyle \frac{{1-\cos \phi }}{2}}}\\\\\therefore \ \ \ \sin \displaystyle \frac{\phi }{2}=\pm \sqrt{{\displaystyle \frac{{1-\displaystyle \frac{3}{5}}}{2}}}\\\\\therefore \ \ \ \sin \displaystyle \frac{\phi }{2}=\pm \displaystyle \frac{{\sqrt{5}}}{5}\\\\\ \ \ \ \ \text{Since}\ \text{270}{}^\circ <\phi <\text{360}{}^\circ ,\\\\\ \ \ \ \ \text{135}{}^\circ <\displaystyle \frac{\phi }{2}<\text{180}{}^\circ .\\\\\therefore \ \ \ \sin \displaystyle \frac{\phi }{2}=\displaystyle \frac{{\sqrt{5}}}{5}\\\\\ \ \ \ \ \sin (\theta +\phi )\\\\\ \ =\ \sin \theta \cos \phi +\cos \theta \sin \phi \\\\\ \ =\ \left( {-\displaystyle \frac{3}{5}} \right)\left( {\displaystyle \frac{3}{5}} \right)+\left( {-\displaystyle \frac{4}{5}} \right)\left( {-\displaystyle \frac{4}{5}} \right)\\\\\ \ =\ -\displaystyle \frac{7}{{25}}\end{array}$

19.    Given that $ \displaystyle \sin \alpha =\frac{5}{13} ,$ where $ \displaystyle 90{}^\circ < \alpha < 180{}^\circ$ and $ \displaystyle \cos \beta = -\frac{3}{5},$ where $ \displaystyle 180{}^\circ < \beta < 360{}^\circ$ find the values of

(a) $ \displaystyle \tan (\alpha + 45{}^\circ) \ \ \ \ \ \ $ (b) $ \displaystyle \sin (\alpha + \beta) \ \ \ \ \ \ $ (c) $ \displaystyle \cos 2\alpha \ \ \ \ \ \ $ (d) $ \displaystyle \sin 2\beta.$

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \sin \alpha =\displaystyle \frac{5}{{13}},\ \ 90{}^\circ <\alpha <180{}^\circ \\\\\therefore \ \ \ \alpha \ \text{is in the }{{\text{2}}^{{\text{nd}}}}\text{ quadrant}\text{.}\end{array}$

$ \displaystyle \therefore \ \ \ \cos \alpha =-\frac{{12}}{{13}},\ \tan \alpha =-\frac{5}{{12}}$


$ \displaystyle \begin{array}{l}\ \ \ \ \ \cos \beta =-\displaystyle \frac{3}{5},\ \ 180{}^\circ <\beta <360{}^\circ \\\\\therefore \ \ \ \beta \ \text{is in the }{{\text{3}}^{{\text{rd}}}}\text{ quadrant}\text{.}\end{array}$

$ \displaystyle \begin{array}{l}\therefore \ \ \ \ \sin \beta =-\displaystyle \frac{4}{5},\ \cos \beta =-\displaystyle \frac{3}{5}\\\\\text{(a)}\ \ \ \tan \left( {\alpha +45{}^\circ } \right)\\\\\ \ \ =\ \displaystyle \frac{{\tan \alpha +\tan 45{}^\circ }}{{1-\tan \alpha \tan 45{}^\circ }}\\\\\ \ \ =\ \displaystyle \frac{{-\displaystyle \frac{5}{{12}}+1}}{{1-\left( {-\displaystyle \frac{5}{{12}}} \right)}}\\\\\ \ \ =\ \displaystyle \frac{{\displaystyle \frac{7}{{12}}}}{{\displaystyle \frac{{17}}{{12}}}}\\\\\ \ \ =\ \displaystyle \frac{7}{{17}}\\\\\text{(b) }\ \ \sin \left( {\alpha +\beta } \right)\\\\\ \ \ =\ \sin \alpha \cos \beta +\cos \alpha \sin \beta \\\\\ \ \ =\ \left( {\displaystyle \frac{5}{{13}}} \right)\left( {-\displaystyle \frac{3}{5}} \right)+\left( {-\displaystyle \frac{{12}}{{13}}} \right)\left( {-\displaystyle \frac{4}{5}} \right)\\\\\ \ \ =\ \displaystyle \frac{{33}}{{65}}\\\\\text{(c)}\ \ \cos 2\alpha \\\\\ \ \ =\ 1-2{{\sin }^{2}}\alpha \\\\\ \ \ =\ 1-2{{\left( {\displaystyle \frac{5}{{13}}} \right)}^{2}}\\\\\ \ \ =\ \displaystyle \frac{{119}}{{169}}\\\\\text{(d)}\ \ \sin 2\beta \\\\\ \ \ =\ 2\sin \beta \cos \beta \\\\\ \ \ =\ 2\left( {-\displaystyle \frac{4}{5}} \right)\left( {-\displaystyle \frac{3}{5}} \right)\\\\\ \ \ =\ \displaystyle \frac{{24}}{{25}}\end{array}$

20.    Given that $ \displaystyle \frac{{\sin (\alpha +\beta )}}{{\sin (\alpha -\beta )}}=\frac{5}{2},$ show that $ \displaystyle 3 \tan \alpha = 7 \tan \beta.$ Given further that $ \displaystyle \alpha + \beta = 45{}^\circ,$ find the value of $ \displaystyle \tan \alpha + \tan \beta.$

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \displaystyle \frac{{\sin \left( {\alpha +\beta } \right)}}{{\sin \left( {\alpha -\beta } \right)}}=\displaystyle \frac{5}{2}\\\\\ \ \ \ \ \ \displaystyle \frac{{\sin \left( {\alpha +\beta } \right)+\sin \left( {\alpha -\beta } \right)}}{{\sin \left( {\alpha +\beta } \right)-\sin \left( {\alpha -\beta } \right)}}=\displaystyle \frac{{5+2}}{{5-2}}\\\\\ \ \ \ \ \ \ \ \left[ \begin{array}{l}\because \displaystyle \frac{a}{b}=\displaystyle \frac{c}{d}\Leftrightarrow \displaystyle \frac{{a+b}}{{a-b}}=\displaystyle \frac{{c+d}}{{c-d}}\\\ (\text{Componendo and Dividendo})\end{array} \right]\\\\\ \therefore \ \ \ \ \displaystyle \frac{{\sin \left( {\alpha +\beta } \right)+\sin \left( {\alpha -\beta } \right)}}{{\sin \left( {\alpha +\beta } \right)-\sin \left( {\alpha -\beta } \right)}}=\displaystyle \frac{7}{2}\\\\\ \therefore \ \ \ \ \displaystyle \frac{{2\sin \displaystyle \frac{{\alpha +\beta +\alpha -\beta }}{2}\cos \displaystyle \frac{{\alpha +\beta -\alpha +\beta }}{2}}}{{2\cos \displaystyle \frac{{\alpha +\beta +\alpha -\beta }}{2}\sin \displaystyle \frac{{\alpha +\beta -\alpha +\beta }}{2}}}=\displaystyle \frac{7}{3}\\\\\ \therefore \ \ \ \ \displaystyle \frac{{2\sin \alpha \cos \beta }}{{2\cos \alpha \sin \beta }}=\displaystyle \frac{7}{3}\\\\\ \therefore \ \ \ \ \tan \alpha \cot \beta =\displaystyle \frac{7}{3}\\\\\ \therefore \ \ \ \ \displaystyle \frac{{\tan \alpha }}{{\tan \beta }}=\displaystyle \frac{7}{3}\\\\\therefore \ \ \ \ \ 3\tan \alpha =7\tan \beta \end{array}$

$ \displaystyle \ \ \ \ \ \ \ \underline{{\text{Solution (1)}\ \text{to find}\tan \alpha +\tan \beta \text{ }}}$

$ \displaystyle \begin{array}{l}\therefore \ \ \ \ \ \tan \beta =\displaystyle \frac{3}{7}\tan \alpha \\\\\ \ \ \ \ \ \ \alpha +\beta =45{}^\circ \\\\\ \ \ \ \ \ \ \tan \left( {\alpha +\beta } \right)=\tan 45{}^\circ \\\\\ \ \ \ \ \ \ \displaystyle \frac{{\tan \alpha +\displaystyle \frac{3}{7}\tan \alpha }}{{1-\tan \alpha \displaystyle \frac{3}{7}\tan \alpha }}=1\ \\\\\therefore \ \ \ \ \ \displaystyle \frac{{10}}{7}\tan \alpha =1-\displaystyle \frac{3}{7}{{\tan }^{2}}\alpha \\\\\therefore \ \ \ \ \ 3{{\tan }^{2}}\alpha +10\tan \alpha -7=0\\\\\therefore \ \ \ \ \ \tan \alpha =\displaystyle \frac{{-10\pm \sqrt{{100-4(3)(-7)}}}}{{2(3)}}\\\\\therefore \ \ \ \ \ \tan \alpha =\displaystyle \frac{{-5\pm \sqrt{{46}}}}{3}\\\\\therefore \ \ \ \ \ \tan \alpha =-3.93\ (\text{or})\ \tan \alpha =0.594\\\\\therefore \ \ \ \ \ \tan \alpha +\tan \beta \\\\\ \ \ \ =\ \tan \alpha +\displaystyle \frac{3}{7}\tan \alpha \\\\\ \ \ \ =\ \displaystyle \frac{{10}}{7}\tan \alpha \\\\\ \ \ \ =\displaystyle \frac{{10}}{7}\times \displaystyle \frac{{-5\pm \sqrt{{46}}}}{3}\\\\\ \ \ \ =\displaystyle \frac{{10\left( {-5\pm \sqrt{{46}}} \right)}}{{21}}\\\\\therefore \ \ \ \ \ \tan \alpha +\tan \beta =-5.61\ \\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\text{or})\\\\\ \ \ \ \ \tan \alpha +\tan \beta =0.8487\ \end{array}$

$ \displaystyle \ \ \ \ \ \ \ \underline{{\text{Solution (2)}\ \text{to find}\tan \alpha +\tan \beta \text{ }}}$

$ \displaystyle \begin{array}{l}\therefore \ \ \ \ \ \displaystyle \frac{{\sin \left( {\alpha +\beta } \right)}}{{\sin \left( {\alpha -\beta } \right)}}=\displaystyle \frac{5}{2}\ \ \ \left[ {\because \text{given}} \right]\\\\\ \ \ \ \ \ \ \text{Let }\sin \left( {\alpha +\beta } \right)=5k\ \text{and}\\\\\ \ \ \ \ \ \ \sin \left( {\alpha -\beta } \right)=2k\\\\\ \ \ \ \ \ \ \text{Since}\ \alpha +\beta =45{}^\circ ,\\\\\ \ \ \ \ \ \ \sin \left( {\alpha +\beta } \right)=\displaystyle \frac{1}{{\sqrt{2}}}\\\\\therefore \ \ \ \ \ \ 5k=\displaystyle \frac{1}{{\sqrt{2}}}\Rightarrow k=\displaystyle \frac{1}{{5\sqrt{2}}}\\\\\therefore \ \ \ \ \ \ \sin \left( {\alpha -\beta } \right)=\displaystyle \frac{2}{{5\sqrt{2}}}=0.2828\\\\\therefore \ \ \ \ \ \ \alpha -\beta =16{}^\circ 2{6}'\ \ \ \text{(or)}\ \ \ \alpha -\beta =180{}^\circ -16{}^\circ 2{6}'\\\\\therefore \ \ \ \ \ \ \alpha -\beta =16{}^\circ 2{6}'\ \ \ \text{(or)}\ \ \ \alpha -\beta =163{}^\circ 3{4}'\\\\\therefore \ \ \ \ \ \ \alpha =30{}^\circ 4{3}',\ \beta =14{}^\circ 1{7}'\ \text{(or)}\ \\\\\ \ \ \ \ \ \ \ \alpha =104{}^\circ 1{7}',\ \beta =-59{}^\circ 1{7}'\\\\\therefore \ \ \ \ \ \tan \alpha +\tan \beta =-5.61\ \\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\text{or})\\\\\ \ \ \ \ \tan \alpha +\tan \beta =0.8487\ \end{array}$

α + β = 45° သည္ α ႏွင့္ β သည္ acute angle ျဖစ္သည္ဟု ဆိုလိုျခင္းမဟုတ္ပါ။ α ႏွင့္ β သည္ acute angle ျဖစ္သည္ဟု ေမးခြန္းတြင္လည္း သတ္မွတ္ထားျခင္း မရွိပါ။ ထိုသို႕မဟုတ္ေၾကာင္းကို solution (2) တြင္ အထင္အရွား ေတြ႔ ႏိုင္ပါသည္။ ထို႕ေၾကာင့္ acute angle တစ္ခုတည္း အတြက္သာ အေျဖထုတ္ျခင္းသည္ မျပည့္စံုပါ။

21.     Express the following as factors.

$ \displaystyle \begin{array}{l}\text{(a)}\sin 5x+\sin 3x\ \ \ \ \text{ (b)}\cos 3x-\cos 5x\\\\\text{(c)}\sin 3x+\sin x\ \ \ \ \text{ }\ \ \text{(d)}\ \sin 5x-\sin x\\\\\text{(e)}\ \cos 2x+\cos 7x\ \ \ \text{ (f)}\cos 9x-\cos x\end{array}$

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$ \displaystyle \begin{array}{l}\text{(a)}\ \ \ \sin 5x+\sin 3x\\\\\ \ \ \ =2\sin \displaystyle \frac{{5x+3x}}{2}\cos \displaystyle \frac{{5x-3x}}{2}\\\\\ \ \ \ =2\sin 4x\cos x\\\\\text{(b)}\ \ \ \cos 3x-\cos 5x\\\\\ \ \ \ =-2\sin \displaystyle \frac{{3x+5x}}{2}\sin \displaystyle \frac{{3x-5x}}{2}\\\\\ \ \ \ =-2\sin 4x\sin (-x)\\\\\ \ \ \ =2\sin 4x\sin x\ \ \ \ \left[ {\because \sin (-x)=-\sin x} \right]\\\\\text{(c)}\ \ \sin 3x+\sin x\\\\\ \ \ \ =2\sin \displaystyle \frac{{3x+x}}{2}\cos \displaystyle \frac{{3x-x}}{2}\\\\\ \ \ \ =2\sin 2x\cos x\\\\\text{(d)}\ \ \sin 5x-\sin x\\\\\ \ \ \ =2\cos \displaystyle \frac{{5x+x}}{2}\sin \displaystyle \frac{{5x-x}}{2}\\\\\ \ \ \ =2\cos 3x\sin 2x\\\\\text{(e)}\ \ \cos 2x+\sin 7x\\\\\ \ \ \ =2\cos \displaystyle \frac{{2x+7x}}{2}\cos \displaystyle \frac{{2x-7x}}{2}\\\\\ \ \ \ =2\cos \displaystyle \frac{{9x}}{2}\cos \left( {- \displaystyle \frac{{5x}}{2}} \right)\\\\\ \ \ \ =2\cos \displaystyle \frac{{9x}}{2}\cos \displaystyle \frac{{5x}}{2}\ \ \left[ {\because \cos (-x)=\cos x} \right]\\\\\text{(f)}\ \ \ \cos 9x-\cos x\\\\\ \ \ \ =-2\sin \displaystyle \frac{{9x+x}}{2}\sin \displaystyle \frac{{9x-x}}{2}\\\\\ \ \ \ =-2\sin 5x\sin 4x\end{array}$

22.     $ \displaystyle \text{If }\alpha +\beta +\gamma =180{}^\circ ,\ \text{prove that}$

$ \displaystyle \begin{array}{l}\text{(a)}\ \tan \alpha +\tan \beta +\tan \gamma =\tan \alpha \tan \beta \tan \gamma \\\ \ \ \ \text{ }\\\text{(b)}\ \sin \alpha +\sin \beta +\sin \gamma =4\cos \displaystyle \frac{\alpha }{2}\cos \displaystyle \frac{\beta }{2}\cos \displaystyle \frac{\gamma }{2}\\\\\text{(c)}\ \sin 2\alpha +\sin 2\beta +\sin 2\gamma =\ 4\sin \alpha \sin \beta \sin \gamma \\\ \text{ }\ \ \\\text{(d)}\ \sin \alpha -\sin \beta +\sin \gamma =4\sin \displaystyle \frac{\alpha }{2}\cos \displaystyle \frac{\beta }{2}\sin \displaystyle \frac{\gamma }{2}\\\\\text{(e)}\ \sin (\alpha +\beta )+\sin (\beta +\gamma )=2\cos \displaystyle \frac{\beta }{2}\cos \displaystyle \frac{{\alpha -\gamma }}{2}\\\\\text{(f)}\ \sin (\alpha +\beta )-\sin (\beta +\gamma )=-2\sin \displaystyle \frac{\beta }{2}\sin \displaystyle \frac{{\alpha -\gamma }}{2}\\\\\text{(g)}\ \cot \displaystyle \frac{\alpha }{2}+\cot \displaystyle \frac{\beta }{2}+\cot \displaystyle \frac{\gamma }{2}=\cot \displaystyle \frac{\alpha }{2}\cot \displaystyle \frac{\beta }{2}\cot \displaystyle \frac{\gamma }{2}\\\\\text{(h)}\ {{\sin }^{2}}\alpha +{{\sin }^{2}}\beta +{{\sin }^{2}}\gamma =2+2\cos \alpha \cos \beta \cos \gamma \\\\\text{(i)}\ {{\sin }^{2}}\alpha +{{\sin }^{2}}\beta -{{\sin }^{2}}\gamma =2\sin \alpha \cos \beta \sin \gamma \\\\\text{(j)}\ {{\cos }^{2}}\alpha +{{\cos }^{2}}\beta +{{\cos }^{2}}\gamma =1-2\cos \alpha \cos \beta \cos \gamma \end{array}$

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$ \displaystyle \begin{array}{l}\text{(a) }\alpha +\beta +\gamma =180\begin{array}{*{20}{l}} {}^\circ \end{array}\\\\\therefore \ \ \ \alpha +\beta =180\begin{array}{*{20}{l}} {}^\circ \end{array}-\gamma \\\\\therefore \ \ \ \displaystyle \frac{{\alpha +\beta }}{2}=90\begin{array}{*{20}{l}} {}^\circ \end{array}-\displaystyle \frac{\gamma }{2}\\\\\therefore \ \ \ \displaystyle \frac{\gamma }{2}=90\begin{array}{*{20}{l}} {}^\circ \end{array}-\displaystyle \frac{{\alpha +\beta }}{2}\\\\\ \ \ \ \ \tan (\alpha +\beta )=\tan (180\begin{array}{*{20}{l}} {}^\circ \end{array}-\gamma )\\\\\therefore \ \ \ \displaystyle \frac{{\tan \alpha +\tan \beta }}{{1-\tan \alpha \tan \beta }}=-\tan \gamma \\\\\therefore \ \ \ \tan \alpha +\tan \beta =-\tan \gamma +\tan \alpha \tan \beta \tan \gamma \\\\\therefore \ \ \ \tan \alpha +\tan \beta +\tan \gamma =\tan \alpha \tan \beta \tan \gamma \end{array}$


$ \displaystyle \begin{array}{l}\text{(b) }\alpha +\beta +\gamma =180\begin{array}{*{20}{l}} {}^\circ \end{array}\\\\\therefore \ \ \ \alpha +\beta =180\begin{array}{*{20}{l}} {}^\circ \end{array}-\gamma \\\\\therefore \ \ \ \gamma =180\begin{array}{*{20}{l}} {}^\circ \end{array}-(\alpha +\beta )\\\\\therefore \ \ \ \displaystyle \frac{{\alpha +\beta }}{2}=90\begin{array}{*{20}{l}} {}^\circ \end{array}-\displaystyle \frac{\gamma }{2}\\\\\therefore \ \ \ \displaystyle \frac{\gamma }{2}=90\begin{array}{*{20}{l}} {}^\circ \end{array}-\displaystyle \frac{{\alpha +\beta }}{2}\\\\\ \ \ \ \ \sin \alpha +\sin \beta +\sin \gamma \\\\=\ \ \left[ {\ \sin \alpha +\sin \beta } \right]+\sin \left[ {180\begin{array}{*{20}{l}} {}^\circ \end{array}-(\alpha +\beta )} \right]\\\\=\ \ 2\sin \displaystyle \frac{{\alpha +\beta }}{2}\cos \displaystyle \frac{{\alpha -\beta }}{2}+\sin \left[ {180\begin{array}{*{20}{l}} {}^\circ \end{array}-(\alpha +\beta )} \right]\\\\=\ \ 2\sin \displaystyle \frac{{\alpha +\beta }}{2}\cos \displaystyle \frac{{\alpha -\beta }}{2}+\sin (\alpha +\beta )\\\\=\ \ 2\sin \displaystyle \frac{{\alpha +\beta }}{2}\cos \displaystyle \frac{{\alpha -\beta }}{2}+\sin 2\left( {\displaystyle \frac{{\alpha +\beta }}{2}} \right)\\\\=\ \ 2\sin \displaystyle \frac{{\alpha +\beta }}{2}\cos \displaystyle \frac{{\alpha -\beta }}{2}+2\sin \displaystyle \frac{{\alpha +\beta }}{2}\cos \displaystyle \frac{{\alpha +\beta }}{2}\\\\=\ \ 2\sin \displaystyle \frac{{\alpha +\beta }}{2}\left[ {\cos \displaystyle \frac{{\alpha +\beta }}{2}+\cos \displaystyle \frac{{\alpha -\beta }}{2}} \right]\\\\=\ \ 2\sin \left( {90\begin{array}{*{20}{l}} {}^\circ \end{array}-\displaystyle \frac{\gamma }{2}} \right)\left[ {2\cos \displaystyle \frac{{\displaystyle \frac{{\alpha +\beta }}{2}+\displaystyle \frac{{\alpha -\beta }}{2}}}{2}\cos \displaystyle \frac{{\displaystyle \frac{{\alpha +\beta }}{2}-\displaystyle \frac{{\alpha -\beta }}{2}}}{2}} \right]\\\\=4\cos \displaystyle \frac{\gamma }{2}\cos \displaystyle \frac{\alpha }{2}\cos \displaystyle \frac{\beta }{2}\\\\\therefore \ \ \ \sin \alpha +\sin \beta +\sin \gamma =4\cos \displaystyle \frac{\alpha }{2}\cos \displaystyle \frac{\beta }{2}\cos \displaystyle \frac{\gamma }{2}\end{array}$


$ \displaystyle \begin{array}{l}\text{(c) }\alpha +\beta +\gamma =180\begin{array}{*{20}{l}} {}^\circ \end{array}\\\\\therefore \ \ \ \alpha +\beta =180\begin{array}{*{20}{l}} {}^\circ \end{array}-\gamma \\\\\therefore \ \ \ \gamma =180\begin{array}{*{20}{l}} {}^\circ \end{array}-(\alpha +\beta )\\\\\ \ \ \ \ \sin 2\alpha +\sin 2\beta +\sin 2\gamma \\\\=\ \ 2\sin \displaystyle \frac{{2\alpha +2\beta }}{2}\cos \displaystyle \frac{{2\alpha -2\beta }}{2}+2\sin \gamma \cos \gamma \\\\=\ \ 2\sin (\alpha +\beta )\cos (\alpha -\beta )+2\sin \gamma \cos \gamma \\\\=\ \ 2\sin (180\begin{array}{*{20}{l}} {}^\circ \end{array}-\gamma )\cos (\alpha -\beta )+2\sin \gamma \cos \left[ {180\begin{array}{*{20}{l}} {}^\circ \end{array}-(\alpha +\beta )} \right]\\\\=\ \ 2\sin \gamma \cos (\alpha -\beta )-2\sin \gamma \cos (\alpha +\beta )\\\\=\ \ 2\sin \gamma \left[ {\cos (\alpha -\beta )-\cos (\alpha +\beta )} \right]\\\\=\ \ 2\sin \gamma \left[ {-2\sin \displaystyle \frac{{(\alpha -\beta )+(\alpha +\beta )}}{2}\sin \displaystyle \frac{{(\alpha -\beta )-(\alpha +\beta )}}{2}} \right]\\\\=\ \ -4\sin \gamma \sin \alpha \sin (-\beta )\\\\=\ \ 4\sin \alpha \sin \beta \sin \gamma \\\\\therefore \ \ \ \sin 2\alpha +\sin 2\beta +\sin 2\gamma =4\sin \alpha \sin \beta \sin \gamma \end{array}$


$ \displaystyle \begin{array}{l}\text{(d) }\alpha +\beta +\gamma =180\begin{array}{*{20}{l}} {}^\circ \end{array}\\\\\therefore \ \ \ \alpha +\beta =180\begin{array}{*{20}{l}} {}^\circ \end{array}-\gamma \\\\\therefore \ \ \ \gamma =180\begin{array}{*{20}{l}} {}^\circ \end{array}-(\alpha +\beta )\\\\\therefore \ \ \ \displaystyle \frac{{\alpha +\beta }}{2}=90\begin{array}{*{20}{l}} {}^\circ \end{array}-\displaystyle \frac{\gamma }{2}\\\\\therefore \ \ \ \displaystyle \frac{\gamma }{2}=90\begin{array}{*{20}{l}} {}^\circ \end{array}-\displaystyle \frac{{\alpha +\beta }}{2}\\\\\ \ \ \ \ \sin \alpha -\sin \beta +\sin \gamma \\\\=\ \ \ \sin \alpha -\sin \beta +\sin \left[ {180\begin{array}{*{20}{l}} {}^\circ \end{array}-(\alpha +\beta )} \right]\\\\=\,\ \ \ \left( {\sin \alpha -\sin \beta } \right)+\sin (\alpha +\beta )\\\\=\,\ \ \ \left( {\sin \alpha -\sin \beta } \right)+\sin 2\left( {\displaystyle \frac{{\alpha +\beta }}{2}} \right)\\\\=\ \ 2\cos \displaystyle \frac{{\alpha +\beta }}{2}\sin \displaystyle \frac{{\alpha -\beta }}{2}+2\sin \displaystyle \frac{{\alpha +\beta }}{2}\cos \displaystyle \frac{{\alpha +\beta }}{2}\\\\=\ \ 2\cos \displaystyle \frac{{\alpha +\beta }}{2}\left[ {\sin \displaystyle \frac{{\alpha +\beta }}{2}+\sin \displaystyle \frac{{\alpha -\beta }}{2}} \right]\\\\=\ \ 2\cos \left( {90\begin{array}{*{20}{l}} {}^\circ \end{array}-\displaystyle \frac{\gamma }{2}} \right)\left[ {2\sin \displaystyle \frac{{\displaystyle \frac{{\alpha +\beta }}{2}+\displaystyle \frac{{\alpha -\beta }}{2}}}{2}\cos \displaystyle \frac{{\displaystyle \frac{{\alpha +\beta }}{2}-\displaystyle \frac{{\alpha -\beta }}{2}}}{2}} \right]\\\\=\ \ 4\sin \displaystyle \frac{\gamma }{2}\sin \displaystyle \frac{\alpha }{2}\cos \displaystyle \frac{\beta }{2}\\\\=\ \ 4\sin \displaystyle \frac{\alpha }{2}\cos \displaystyle \frac{\beta }{2}\sin \displaystyle \frac{\gamma }{2}\\\\\therefore \ \ \sin \alpha -\sin \beta +\sin \gamma =4\sin \displaystyle \frac{\alpha }{2}\cos \displaystyle \frac{\beta }{2}\sin \displaystyle \frac{\gamma }{2}\end{array}$


$ \displaystyle \begin{array}{l}\text{(e) }\alpha +\beta +\gamma =180\begin{array}{*{20}{l}} {}^\circ \end{array}\\\\\ \ \ \ \ \sin (\alpha +\beta )+\sin (\beta +\gamma )\\\\=\ \ \ 2\sin \displaystyle \frac{{\alpha +\beta +\beta +\gamma }}{2}\cos \displaystyle \frac{{\alpha +\beta -\beta -\gamma }}{2}\\\\=\ \ \ 2\sin \displaystyle \frac{{180\begin{array}{*{20}{l}} {}^\circ \end{array}+\beta }}{2}\cos \displaystyle \frac{{\alpha -\gamma }}{2}\\\\=\ \ \ 2\sin \left( {90\begin{array}{*{20}{l}} {}^\circ \end{array}+\displaystyle \frac{\beta }{2}} \right)\cos \displaystyle \frac{{\alpha -\gamma }}{2}\\\\=\ \ \ 2\cos \displaystyle \frac{\beta }{2}\cos \displaystyle \frac{{\alpha -\gamma }}{2}\\\\\therefore \ \ \ \sin (\alpha +\beta )+\sin (\beta +\gamma )=2\cos \displaystyle \frac{\beta }{2}\cos \displaystyle \frac{{\alpha -\gamma }}{2}\end{array}$


$ \displaystyle \begin{array}{l}\text{(f) }\alpha +\beta +\gamma =180\begin{array}{*{20}{l}} {}^\circ \end{array}\\\\\ \ \ \ \ \sin (\alpha +\beta )-\sin (\beta +\gamma )\\\\=\ \ \ 2\cos \displaystyle \frac{{\alpha +\beta +\beta +\gamma }}{2}\sin \displaystyle \frac{{\alpha +\beta -\beta -\gamma }}{2}\\\\=\ \ \ 2\cos \displaystyle \frac{{180\begin{array}{*{20}{l}} {}^\circ \end{array}+\beta }}{2}\sin \displaystyle \frac{{\alpha -\gamma }}{2}\\\\=\ \ \ 2\cos \left( {90\begin{array}{*{20}{l}} {}^\circ \end{array}+\displaystyle \frac{\beta }{2}} \right)\sin \displaystyle \frac{{\alpha -\gamma }}{2}\\\\=\ \ \ -2\sin \displaystyle \frac{\beta }{2}\sin \displaystyle \frac{{\alpha -\gamma }}{2}\\\\\therefore \ \ \ \sin (\alpha +\beta )-\sin (\beta +\gamma )=-2\sin \displaystyle \frac{\beta }{2}\sin \displaystyle \frac{{\alpha -\gamma }}{2}\end{array}$


$ \displaystyle \begin{array}{l}\text{(g) }\alpha +\beta +\gamma =180\begin{array}{*{20}{l}} {}^\circ \end{array}\\\\\therefore \ \ \ \alpha +\beta =180\begin{array}{*{20}{l}} {}^\circ \end{array}-\gamma \\\\\therefore \ \ \ \displaystyle \frac{\alpha }{2}+\displaystyle \frac{\beta }{2}=90\begin{array}{*{20}{l}} {}^\circ \end{array}-\displaystyle \frac{\gamma }{2}\\\\\therefore \ \ \ \tan \left( {\displaystyle \frac{\alpha }{2}+\displaystyle \frac{\beta }{2}} \right)=\tan \left( {90\begin{array}{*{20}{l}} {}^\circ \end{array}-\displaystyle \frac{\gamma }{2}} \right)\\\\\therefore \ \ \ \displaystyle \frac{{\tan \displaystyle \frac{\alpha }{2}+\tan \displaystyle \frac{\beta }{2}}}{{1-\tan \displaystyle \frac{\alpha }{2}\tan \displaystyle \frac{\beta }{2}}}=\cot \displaystyle \frac{\gamma }{2}\\\\\therefore \ \ \ \displaystyle \frac{{\displaystyle \frac{1}{{\cot \displaystyle \frac{\alpha }{2}}}+\displaystyle \frac{1}{{\cot \displaystyle \frac{\beta }{2}}}}}{{1-\displaystyle \frac{1}{{\cot \displaystyle \frac{\alpha }{2}\cot \displaystyle \frac{\beta }{2}}}}}=\cot \displaystyle \frac{\gamma }{2}\\\\\therefore \ \ \ \displaystyle \frac{{\displaystyle \frac{{\cot \displaystyle \frac{\alpha }{2}+\cot \displaystyle \frac{\beta }{2}}}{{\cot \displaystyle \frac{\alpha }{2}\cot \displaystyle \frac{\beta }{2}}}}}{{\displaystyle \frac{{\cot \displaystyle \frac{\alpha }{2}\cot \displaystyle \frac{\beta }{2}-1}}{{\cot \displaystyle \frac{\alpha }{2}\cot \displaystyle \frac{\beta }{2}}}}}=\cot \displaystyle \frac{\gamma }{2}\\\\\therefore \ \ \ \displaystyle \frac{{\cot \displaystyle \frac{\alpha }{2}+\cot \displaystyle \frac{\beta }{2}}}{{\cot \displaystyle \frac{\alpha }{2}\cot \displaystyle \frac{\beta }{2}-1}}=\cot \displaystyle \frac{\gamma }{2}\\\\\therefore \ \ \ \cot \displaystyle \frac{\alpha }{2}+\cot \displaystyle \frac{\beta }{2}=\cot \displaystyle \frac{\alpha }{2}\cot \displaystyle \frac{\beta }{2}\cot \displaystyle \frac{\gamma }{2}-\cot \displaystyle \frac{\gamma }{2}\\\\\therefore \ \ \ \cot \displaystyle \frac{\alpha }{2}+\cot \displaystyle \frac{\beta }{2}+\cot \displaystyle \frac{\gamma }{2}=\cot \displaystyle \frac{\alpha }{2}\cot \displaystyle \frac{\beta }{2}\cot \displaystyle \frac{\gamma }{2}\end{array}$


$ \displaystyle \begin{array}{l}\text{(h) }\alpha +\beta +\gamma =180\begin{array}{*{20}{l}} {}^\circ \end{array}\\\\\therefore \ \ \ \alpha +\beta =180\begin{array}{*{20}{l}} {}^\circ \end{array}-\gamma \\\\\therefore \ \ \ \gamma =180\begin{array}{*{20}{l}} {}^\circ \end{array}-(\alpha +\beta )\\\\\ \ \ \ \ {{\sin }^{2}}\alpha +{{\sin }^{2}}\beta +{{\sin }^{2}}\gamma \\\\=\ \ \ \displaystyle \frac{1}{2}(2{{\sin }^{2}}\alpha +2{{\sin }^{2}}\beta )+1-{{\cos }^{2}}\gamma \\\\=\ \ \ \displaystyle \frac{1}{2}(1-1+2{{\sin }^{2}}\alpha +1-1+2{{\sin }^{2}}\beta )+1-{{\cos }^{2}}\gamma \\\\=\ \ \ \displaystyle \frac{1}{2}\left[ {1-\left( {1-2{{{\sin }}^{2}}\alpha } \right)+1-\left( {1-2{{{\sin }}^{2}}\beta } \right)} \right]+1-{{\cos }^{2}}\gamma \\\\=\ \ \ \displaystyle \frac{1}{2}\left[ {2-\cos 2\alpha -\cos 2\beta } \right]+1-{{\cos }^{2}}\gamma \\\\=\ \ \ 1-\displaystyle \frac{1}{2}\left( {\cos 2\alpha +\cos 2\beta } \right)+1-{{\cos }^{2}}\gamma \\\\=\ \ \ 1-\displaystyle \frac{1}{2}\left( {2\cos \displaystyle \frac{{2\alpha +2\beta }}{2}\cos \displaystyle \frac{{2\alpha -2\beta }}{2}} \right)+1-{{\cos }^{2}}\gamma \\\\=\ \ \ 1-\cos \left( {\alpha +\beta } \right)\cos \left( {\alpha -\beta } \right)+1-{{\cos }^{2}}\gamma \\\\=\ \ \ 2-\cos \left( {180\begin{array}{*{20}{l}} {}^\circ \end{array}-\gamma } \right)\cos \left( {\alpha -\beta } \right)-{{\cos }^{2}}\gamma \\\\=\ \ \ 2+\cos \gamma \cos \left( {\alpha -\beta } \right)-{{\cos }^{2}}\gamma \\\\=\ \ \ 2+\cos \gamma \left[ {\cos \left( {\alpha -\beta } \right)-\cos \gamma } \right]\\\\=\ \ \ 2+\cos \gamma \left[ {\cos \left( {\alpha -\beta } \right)-\cos \left( {180\begin{array}{*{20}{l}} {}^\circ \end{array}-(\alpha +\beta )} \right)} \right]\\\\=\ \ \ 2+\cos \gamma \left[ {\cos \left( {\alpha -\beta } \right)+\cos \left( {\alpha +\beta } \right)} \right]\\\\=\ \ \ 2+\cos \gamma \left[ {\cos \left( {\alpha +\beta } \right)+\cos \left( {\alpha -\beta } \right)} \right]\\\\=\ \ \ 2+\cos \gamma \left[ {2\cos \displaystyle \frac{{\alpha +\beta +\alpha -\beta }}{2}\cos \displaystyle \frac{{\alpha +\beta -\alpha +\beta }}{2}} \right]\\\\=\ \ \ 2+2\cos \gamma \cos \alpha \cos \beta \\\\\therefore \ \ \ \ \ \ \ \ {{\sin }^{2}}\alpha +{{\sin }^{2}}\beta +{{\sin }^{2}}\gamma =2+2\cos \alpha \cos \beta \cos \gamma \end{array}$


$ \displaystyle \begin{array}{l}\text{(i) }\alpha +\beta +\gamma =180\begin{array}{*{20}{l}} {}^\circ \end{array}\\\\ \therefore \ \ \ \alpha +\beta =180\begin{array}{*{20}{l}} {}^\circ \end{array}-\gamma \\\\ \therefore \ \ \ \gamma =180\begin{array}{*{20}{l}} {}^\circ \end{array}-(\alpha +\beta )\\\\ \ \ \ \ \ {{\sin }^{2}}\alpha +{{\sin }^{2}}\beta -{{\sin }^{2}}\gamma \\\\ =\ \ \ \displaystyle \frac{1}{2}\left( {2{{{\sin }}^{2}}\alpha +2{{{\sin }}^{2}}\beta } \right)-\left( {1-{{{\cos }}^{2}}\gamma } \right)\\\\ =\ \ \ \displaystyle \frac{1}{2}\left( {1-1+2{{{\sin }}^{2}}\alpha +1-1+2{{{\sin }}^{2}}\beta } \right)-1+{{\cos }^{2}}\gamma \\\\ =\ \ \ \displaystyle \frac{1}{2}\left[ {1-\left( {1-2{{{\sin }}^{2}}\alpha } \right)+1-\left( {1-2{{{\sin }}^{2}}\beta } \right)} \right]-1+{{\cos }^{2}}\gamma \\\\ =\ \ \ \displaystyle \frac{1}{2}\left[ {2-\cos 2\alpha -\cos 2\beta } \right]-1+{{\cos }^{2}}\gamma \\\\ =\ \ \ 1-\displaystyle \frac{1}{2}\left( {\cos 2\alpha +\cos 2\beta } \right)-1+{{\cos }^{2}}\gamma \\\\ =\ \ \ -\displaystyle \frac{1}{2}\left( {2\cos \displaystyle \frac{{2\alpha +2\beta }}{2}\cos \displaystyle \frac{{2\alpha -2\beta }}{2}} \right)+{{\cos }^{2}}\gamma \\\\ =\ \ \ -\cos \left( {\alpha +\beta } \right)\cos \left( {\alpha -\beta } \right)+{{\cos }^{2}}\gamma \\\\ =\ \ \ -\cos \left( {180\begin{array}{*{20}{l}} {}^\circ \end{array}-\gamma } \right)\cos \left( {\alpha -\beta } \right)+{{\cos }^{2}}\gamma \\\\ =\ \ \ \cos \gamma \cos \left( {\alpha -\beta } \right)+{{\cos }^{2}}\gamma \\\\ =\ \ \ \cos \gamma \left[ {\cos \left( {\alpha -\beta } \right)+\cos \gamma } \right]\\\\ =\ \ \cos \gamma \left[ {\cos \left( {\alpha -\beta } \right)+\cos \left( {180\begin{array}{*{20}{l}} {}^\circ \end{array}-(\alpha +\beta )} \right)} \right]\\\\ =\ \ \cos \gamma \left[ {\cos \left( {\alpha -\beta } \right)-\cos \left( {\alpha +\beta } \right)} \right]\\\\ =\ \ \ \cos \gamma \left[ {-2\sin \displaystyle \frac{{\alpha -\beta +\alpha +\beta }}{2}\sin \displaystyle \frac{{\alpha -\beta -\alpha -\beta }}{2}} \right]\\\\ =\ \ \ -2\cos \gamma \sin \alpha \sin \left( {-\beta } \right)\\\\ =\ \ \ 2\sin \alpha \sin \beta \cos \gamma \\\\ \therefore \ \ \ {{\sin }^{2}}\alpha +{{\sin }^{2}}\beta -{{\sin }^{2}}\gamma =2\sin \alpha \sin \beta \cos \gamma \end{array}$


$ \displaystyle \begin{array}{l}\text{(j) }\alpha +\beta +\gamma =180\begin{array}{*{20}{l}} {}^\circ \end{array}\\\\\therefore \ \ \ \alpha +\beta =180\begin{array}{*{20}{l}} {}^\circ \end{array}-\gamma \\\\\therefore \ \ \ \gamma =180\begin{array}{*{20}{l}} {}^\circ \end{array}-(\alpha +\beta )\\\\\ \ \ \ \ {{\cos }^{2}}\alpha +{{\cos }^{2}}\beta +{{\cos }^{2}}\gamma \\\\=\ \ \ \displaystyle \frac{1}{2}\left( {2{{{\cos }}^{2}}\alpha +2{{{\cos }}^{2}}\beta } \right)+{{\cos }^{2}}\gamma \\\\=\ \ \ \displaystyle \frac{1}{2}\left( {1+2{{{\cos }}^{2}}\alpha -1+1+2{{{\cos }}^{2}}\beta -1} \right)+{{\cos }^{2}}\gamma \\\\=\ \ \ \displaystyle \frac{1}{2}\left[ {2+\left( {2{{{\cos }}^{2}}\alpha -1} \right)+\left( {2{{{\cos }}^{2}}\beta -1} \right)} \right]+{{\cos }^{2}}\gamma \\\\=\ \ \ \displaystyle \frac{1}{2}\left[ {2+\cos 2\alpha +\cos 2\beta } \right]+{{\cos }^{2}}\gamma \\\\=\ \ \ 1+\displaystyle \frac{1}{2}\left( {\cos 2\alpha +\cos 2\beta } \right)+{{\cos }^{2}}\gamma \\\\=\ \ \ 1+\displaystyle \frac{1}{2}\left( {2\cos \displaystyle \frac{{2\alpha +2\beta }}{2}\cos \displaystyle \frac{{2\alpha -2\beta }}{2}} \right)+{{\cos }^{2}}\gamma \\\\=\ \ \ 1+\cos \left( {\alpha +\beta } \right)\cos \left( {\alpha -\beta } \right)+{{\cos }^{2}}\gamma \\\\=\ \ \ 1+\cos \left( {180\begin{array}{*{20}{l}} {}^\circ \end{array}-\gamma } \right)\cos \left( {\alpha -\beta } \right)+{{\cos }^{2}}\gamma \\\\=\ \ \ 1-\cos \gamma \cos \left( {\alpha -\beta } \right)+{{\cos }^{2}}\gamma \\\\=\ \ 1-\ \cos \gamma \left[ {\cos \left( {\alpha -\beta } \right)-\cos \gamma } \right]\\\\=\ \ 1-\cos \gamma \left[ {\cos \left( {\alpha -\beta } \right)-\cos \left( {180\begin{array}{*{20}{l}} {}^\circ \end{array}-(\alpha +\beta )} \right)} \right]\\\\=\ \ 1-\cos \gamma \left[ {\cos \left( {\alpha -\beta } \right)+\cos \left( {\alpha +\beta } \right)} \right]\\\\=\ \ 1-\cos \gamma \left[ {2\cos \displaystyle \frac{{\alpha -\beta +\alpha +\beta }}{2}\cos \displaystyle \frac{{\alpha -\beta -\alpha -\beta }}{2}} \right]\\\\=\ \ \ 1-2\cos \gamma \cos \alpha \cos \left( {-\beta } \right)\\\\=\ \ \ 1-2\cos \alpha \cos \beta \cos \gamma \\\\\therefore \ \ \ {{\cos }^{2}}\alpha +{{\cos }^{2}}\beta +{{\cos }^{2}}\gamma =1-2\cos \alpha \cos \beta \cos \gamma \end{array}$

23.     $ \displaystyle \text{If }\alpha +\beta +\gamma =90{}^\circ ,\ \text{prove that}$

$ \displaystyle \begin{array}{l}\text{(a)}\ \sin 2\alpha +\sin 2\beta +\sin 2\gamma =\ 4\cos \alpha \cos \beta \cos \gamma \\\ \text{ }\ \ \\\text{(b)}\ \tan \alpha \tan \beta +\tan \beta \tan \gamma +\tan \gamma \tan \alpha =\text{ }1\end{array}$

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$ \displaystyle \begin{array}{l}\text{(a) }\alpha +\beta +\gamma =90\begin{array}{*{20}{l}} {}^\circ \end{array}\\\\\therefore \ \ \ \alpha +\beta =90\begin{array}{*{20}{l}} {}^\circ \end{array}-\gamma \\\\\therefore \ \ \ \gamma =90\begin{array}{*{20}{l}} {}^\circ \end{array}-(\alpha +\beta )\\\\\ \ \ \ \ \sin 2\alpha +\sin 2\beta +\sin 2\gamma \\\\=\ \ \ 2\sin \displaystyle \frac{{2\alpha +2\beta }}{2}\cos \displaystyle \frac{{2\alpha -2\beta }}{2}+2\sin \gamma \cos \gamma \\\\=\ \ \ 2\sin \left( {\alpha +\beta } \right)\cos \left( {\alpha -\beta } \right)+2\sin \gamma \cos \gamma \\\\=\ \ \ 2\sin \left( {90\begin{array}{*{20}{l}} {}^\circ \end{array}-\gamma } \right)\cos \left( {\alpha -\beta } \right)+2\sin \left[ {90\begin{array}{*{20}{l}} {}^\circ \end{array}-(\alpha +\beta )} \right]\cos \gamma \\\\=\ \ \ 2\cos \gamma \cos \left( {\alpha -\beta } \right)+2\cos \gamma \cos (\alpha +\beta )\\\\=\ \ \ 2\cos \gamma \left[ {\cos \left( {\alpha +\beta } \right)+\cos (\alpha -\beta )} \right]\\\\=\ \ \ 2\cos \gamma \left[ {2\cos \displaystyle \frac{{\alpha +\beta +\alpha -\beta }}{2}\cos \displaystyle \frac{{\alpha +\beta -\alpha +\beta }}{2}} \right]\\\\=4\cos \alpha \cos \beta \cos \gamma \end{array}$


$ \displaystyle \begin{array}{l}\text{(b) }\alpha +\beta +\gamma =90\begin{array}{*{20}{l}} {}^\circ \end{array}\\\\\therefore \ \ \ \alpha +\beta =90\begin{array}{*{20}{l}} {}^\circ \end{array}-\gamma \\\\\therefore \ \ \ \tan \left( {\alpha +\beta } \right)=\tan \left( {90\begin{array}{*{20}{l}} {}^\circ \end{array}-\gamma } \right)\\\\\therefore \ \ \ \displaystyle \frac{{\tan \alpha +\tan \beta }}{{1-\tan \alpha \tan \beta }}=\cot \gamma \\\\\therefore \ \ \ \displaystyle \frac{{\tan \alpha +\tan \beta }}{{1-\tan \alpha \tan \beta }}=\displaystyle \frac{1}{{\tan \gamma }}\\\\\ \therefore \ \ \ \tan \alpha \tan \gamma +\tan \beta \tan \gamma =\ 1-\tan \alpha \tan \beta \\\\\ \therefore \ \ \ \tan \alpha \tan \beta +\tan \beta \tan \gamma +\tan \gamma \tan \alpha =\ 1\end{array}$

24.     If $ \displaystyle \alpha + \beta = \frac{\pi}{4},$ prove that $ \displaystyle (1 + \tan \alpha) (1 + \tan \beta) = 2.$

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$ \displaystyle \begin{array}{l}\ \ \ \ \alpha +\beta =\displaystyle \frac{\pi }{4}\\\\\therefore \ \ \tan (\alpha +\beta )=\tan \displaystyle \frac{\pi }{4}\\\\\therefore \ \ \displaystyle \frac{{\tan \alpha +\tan \beta }}{{1-\tan \alpha \tan \beta }}=1\\\\\therefore \ \ \tan \alpha +\tan \beta =1-\tan \alpha \tan \beta \\\\\therefore \ \ \tan \alpha +\tan \alpha \tan \beta +\tan \beta =1\\\\\therefore \ \ \tan \alpha +\tan \alpha \tan \beta +1+\tan \beta =2\\\\\therefore \ \ \tan \alpha (1+\tan \beta )+(1+\tan \beta )=2\\\\\therefore \ \ (1+\tan \alpha )(1+\tan \beta )=2\end{array}$

25.     If $ \displaystyle \alpha + \beta = \frac{\pi}{2},$ and $ \displaystyle \beta + \gamma = \alpha,$ show that $ \displaystyle \tan \alpha = 2 \tan γ + \tan \beta.$

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$ \displaystyle \begin{array}{l}\ \ \ \ \alpha +\beta =\displaystyle \frac{\pi }{2}\Rightarrow \alpha =\displaystyle \frac{\pi }{2}-\beta \\\\\therefore \ \ \tan \alpha =\tan \left( {\displaystyle \frac{\pi }{2}-\beta } \right)=\cot \beta \\\\\therefore \ \ \tan \alpha =\displaystyle \frac{1}{{\tan \beta }}\\\\\therefore \ \ \tan \alpha \tan \beta =1\\\\\ \ \ \ \beta +\gamma =\alpha \Rightarrow \gamma =\alpha -\beta \\\\\therefore \ \ \tan \gamma =\tan \left( {\alpha -\beta } \right)\ \\\\\therefore \ \ \tan \gamma =\displaystyle \frac{{\tan \alpha -\tan \beta }}{{1+\tan \alpha \tan \beta }}\\\\\therefore \ \ \tan \gamma =\displaystyle \frac{{\tan \alpha -\tan \beta }}{{1+1}}\ \ \ \left[ {\because \tan \alpha \tan \beta =1} \right]\\\\\therefore \ \ 2\tan \gamma =\tan \alpha -\tan \beta \ \\\\\therefore \ \ \tan \alpha =2\tan \gamma +\tan \beta \end{array}$

26.     Prove the following identities:

$ \displaystyle \begin{array}{l}\text{(a)}\ \ \cos (60{}^\circ +x)+\sin (30{}^\circ +x)=\cos x\\\\\text{(b)}\ \ \tan \alpha +\cot \alpha =2\operatorname{cosec}2\alpha \\\\\text{(c)}\ \ \operatorname{cosec}2x+\cot 2x=\cot x\\\\\text{(d)}\ \ \tan 2x\left( {2\cos x\sec x} \right)=2\sin x\\\\\text{(e)}\ \ \left( {\tan x+\sin x} \right)\left( {\tan x-\sin x} \right)\text{=}{{\tan }^{2}}x\ {{\sin }^{2}}x\\\\\text{(f)}\ \ \left( {1+\tan x-\sec x} \right)\left( {1+\cot x-\operatorname{cosec}x} \right)=2\\\\\text{(g)}\ \ \sin \left( {x+y} \right)\sin \left( {x-y} \right)={{\sin }^{2}}x-{{\sin }^{2}}y\\\\\text{(h)}\ \ \displaystyle \frac{{\tan y-\tan x}}{{\tan y+\tan x}}=\displaystyle \frac{{\sin (y-x)}}{{\sin (y+x)}}\\\\\text{(i)}\ \ \displaystyle \frac{{\cos \theta -\cos \phi }}{{\sin \theta +\sin \phi }}=\tan \displaystyle \frac{{\phi -\theta }}{2}\\\\\text{(j)}\ \ \displaystyle \frac{{\sin \alpha +\sin \beta }}{{\cos \alpha -\cos \beta }}=\cot \displaystyle \frac{{\beta -\alpha }}{2}\\\\\text{(k)}\ \ \displaystyle \frac{{\sin 2x}}{{1+\cos 2x}}=\tan x\\\\\text{(l)}\ \ \displaystyle \frac{{\sin 3\alpha }}{{\cos \alpha }}+\displaystyle \frac{{\cos 3\alpha }}{{\sin \alpha }}=2\cot 2\alpha \\\\\text{(m)}\ \displaystyle \frac{1}{{\cos 2\alpha }}-\displaystyle \frac{{\cos 2\alpha }}{{1+\sin 2\alpha }}=\tan 2\alpha \\\\\text{(n)}\ \ \displaystyle \frac{{\tan \alpha }}{{\sec \alpha -1}}+\displaystyle \frac{{\tan \alpha }}{{\sec \alpha +1}}=2\operatorname{cosec}\alpha \\\\\text{(o)}\ \ \displaystyle \frac{{2-{{{\operatorname{cosec}}}^{2}}\alpha }}{{{{{\operatorname{cosec}}}^{2}}\alpha +2\cot \alpha }}=\displaystyle \frac{{\sin \alpha -\cos \alpha }}{{\sin \alpha +\cos \alpha }}\\\\\text{(p)}\ \ 1-\displaystyle \frac{{{{{\sin }}^{2}}x}}{{1+\cos x}}=\cos x\\\\\text{(q)}\ \ \displaystyle \frac{{1+{{{\cot }}^{2}}x}}{{1-{{{\cot }}^{2}}x}}=-\sec 2x\\\\\text{(r)}\ \ \displaystyle \frac{{\cos \alpha +\sec \beta }}{{\cos \beta +\sec \alpha }}=\cos \alpha \sec \beta\end{array}$

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$ \displaystyle \begin{array}{l}\text{(a)}\ \ \ \ \cos \left( {60{}^\circ +x} \right)+\sin \left( {30{}^\circ +x} \right)\\\ \ \ \ \ =\ \cos 60{}^\circ \cos x-\sin 60{}^\circ \sin x+\sin 30{}^\circ \cos x+\cos 30{}^\circ \sin x\\\ \ \ \ \ =\ \displaystyle \frac{1}{2}\cos x-\displaystyle \frac{{\sqrt{3}}}{2}\sin x+\displaystyle \frac{1}{2}\cos x+\displaystyle \frac{{\sqrt{3}}}{2}\sin x\\\ \ \ \ \ =\ \cos x\end{array}$


$ \displaystyle \begin{array}{l}\text{(b)}\ \ \ \ \tan \alpha +\cot \alpha \\\\\ \ \ \ =\ \ \displaystyle \frac{{\sin \alpha }}{{\cos \alpha }}+\displaystyle \frac{{\cos \alpha }}{{\sin \alpha }}\\\\\ \ \ \ =\ \ \displaystyle \frac{{{{{\sin }}^{2}}\alpha +{{{\cos }}^{2}}\alpha }}{{\sin \alpha \cos \alpha }}\\\\\ \ \ \ =\ \ \displaystyle \frac{1}{{\sin \alpha \cos \alpha }}\\\\\ \ \ \ =\ \ \displaystyle \frac{2}{{2\sin \alpha \cos \alpha }}\\\\\ \ \ \ =\ \ \displaystyle \frac{2}{{\sin 2\alpha }}\\\\\ \ \ \ =\ \ 2\operatorname{cosec}2\alpha \end{array}$


$ \displaystyle \begin{array}{l}\text{(c)}\ \ \ \ \operatorname{cosec}2x+\cot 2x\\\\\ \ \ \ =\ \ \displaystyle \frac{1}{{\sin 2x}}+\displaystyle \frac{{\cos 2x}}{{\sin 2x}}\\\\\ \ \ \ =\ \ \displaystyle \frac{{1+\cos 2x}}{{\sin 2x}}\\\\\ \ \ \ =\ \ \displaystyle \frac{{1+2{{{\cos }}^{2}}x-1}}{{2\sin x\cos x}}\\\\\ \ \ \ =\ \ \displaystyle \frac{{{{{\cos }}^{2}}x}}{{\sin x\cos x}}\\\\\ \ \ \ =\ \ \displaystyle \frac{{\cos x}}{{\sin x}}\\\\\ \ \ \ =\ \ \cot x\end{array}$


$ \displaystyle \begin{array}{l}\text{(d)}\ \ \ \ \tan 2x\left( {2\cos x-\sec x} \right)\\\\\ \ \ \ =\ \ \displaystyle \frac{{\sin 2x}}{{\cos 2x}}\left( {2\cos x-\displaystyle \frac{1}{{\cos x}}} \right)\\\\\ \ \ \ =\ \ \displaystyle \frac{{2\sin x\cos x}}{{2{{{\cos }}^{2}}x-1}}\times \displaystyle \frac{{2{{{\cos }}^{2}}x-1}}{{\cos x}}\\\\\ \ \ \ =\ \ 2\sin x\end{array}$


$ \displaystyle \begin{array}{l}\text{(e)}\ \ \ \ \left( {\tan x+\sin x} \right)\left( {\tan x-\sin x} \right)\\\\\ \ \ \ =\ \ {{\tan }^{2}}x-{{\sin }^{2}}x\\\\\ \ \ \ =\ \ \displaystyle \frac{{{{{\sin }}^{2}}x}}{{{{{\cos }}^{2}}x}}-{{\sin }^{2}}x\\\\\ \ \ \ =\ \ \displaystyle \frac{{{{{\sin }}^{2}}x-{{{\sin }}^{2}}x{{{\cos }}^{2}}x}}{{{{{\cos }}^{2}}x}}\\\\\ \ \ \ =\ \ \displaystyle \frac{{{{{\sin }}^{2}}x\left( {1-{{{\cos }}^{2}}x} \right)}}{{{{{\cos }}^{2}}x}}\\\\\ \ \ \ =\ \displaystyle \frac{{{{{\sin }}^{2}}x}}{{{{{\cos }}^{2}}x}}\ {{\sin }^{2}}x\\\\\ \ \ \ =\ {{\tan }^{2}}x\ {{\sin }^{2}}x\end{array}$


$ \displaystyle \begin{array}{l}\text{(f)}\ \ \ \ \left( {1+\tan x-\sec x} \right)\left( {1+\cot x+\operatorname{cosec}x} \right)\\\\\ \ \ \ =\ \ \left( {1+\displaystyle \frac{{\sin x}}{{\cos x}}-\displaystyle \frac{1}{{\cos x}}} \right)\left( {1+\displaystyle \frac{{\cos x}}{{\sin x}}+\displaystyle \frac{1}{{\sin x}}} \right)\\\\\ \ \ \ =\ \ \displaystyle \frac{{\cos x+\sin x-1}}{{\cos x}}\times \displaystyle \frac{{\sin x+\cos x+1}}{{\sin x}}\\\\\ \ \ \ =\ \ \displaystyle \frac{{{{{\left( {\sin x+\cos x} \right)}}^{2}}-1}}{{\sin x\cos x}}\\\\\ \ \ \ =\ \ \displaystyle \frac{{{{{\sin }}^{2}}x+2\sin x\cos x+{{{\cos }}^{2}}x-1}}{{\sin x\cos x}}\\\\\ \ \ \ =\ \ \displaystyle \frac{{1+2\sin x\cos x-1}}{{\sin x\cos x}}\\\\\ \ \ \ =\ \ \displaystyle \frac{{2\sin x\cos x}}{{\sin x\cos x}}\\\\\ \ \ \,=\ \ 2\end{array}$


$ \displaystyle \begin{array}{l}\text{(g)}\ \ \ \ \sin \left( {x+y} \right)\sin \left( {x-y} \right)\\\\\ \ \ \ =\ \ -\displaystyle \frac{1}{2}\left[ {-2\sin \displaystyle \frac{{2x+2y}}{2}\sin \displaystyle \frac{{2x-2y}}{2}} \right]\\\\\ \ \ \ =\ \ -\displaystyle \frac{1}{2}\left( {\cos 2x-\cos 2y} \right)\\\\\ \ \ \ =\ \ -\displaystyle \frac{1}{2}\left( {1-2{{{\sin }}^{2}}x-1+2{{{\sin }}^{2}}y} \right)\\\\\ \ \ \ =\ \ -\displaystyle \frac{1}{2}\left( {-2{{{\sin }}^{2}}x+2{{{\sin }}^{2}}y} \right)\\\\\ \ \ \ =\ \ {{\sin }^{2}}x-{{\sin }^{2}}y\end{array}$


$ \displaystyle \begin{array}{l}\text{(h)}\ \ \ \ \displaystyle \frac{{\tan y-\tan x}}{{\tan y+\tan x}}\\\\\ \ \ \ =\ \ \displaystyle \frac{{\displaystyle \frac{{\sin y}}{{\cos y}}-\displaystyle \frac{{\sin x}}{{\cos x}}}}{{\displaystyle \frac{{\sin y}}{{\cos y}}+\displaystyle \frac{{\sin x}}{{\cos x}}}}\\\\\ \ \ \ =\ \ \displaystyle \frac{{\displaystyle \frac{{\sin y\cos x-\cos y\sin x}}{{\cos y\cos x}}}}{{\displaystyle \frac{{\sin y\cos x+\cos y\sin x}}{{\cos y\cos x}}}}\\\\\ \ \ \ =\ \ \displaystyle \frac{{\sin y\cos x-\cos y\sin x}}{{\sin y\cos x+\cos y\sin x}}\\\\\ \ \ \ =\ \ \displaystyle \frac{{\sin (y-x)}}{{\sin (y+x)}}\end{array}$


$ \displaystyle \begin{array}{l}\text{(i)}\ \ \ \ \displaystyle \frac{{\cos \theta -\cos \phi }}{{\sin \theta +\sin \phi }}\\\\\ \ \ \ =\ \ \displaystyle \frac{{-2\sin \displaystyle \frac{{\theta +\phi }}{2}\sin \displaystyle \frac{{\theta -\phi }}{2}}}{{2\sin \displaystyle \frac{{\theta +\phi }}{2}\cos \displaystyle \frac{{\theta -\phi }}{2}}}\\\\\ \ \ \ =\ \ \displaystyle \frac{{-\sin \displaystyle \frac{{\theta -\phi }}{2}}}{{\cos \displaystyle \frac{{\theta -\phi }}{2}}}\\\\\ \ \ \ =\ \ -\tan \displaystyle \frac{{\theta -\phi }}{2}\\\\\ \ \ \ =\ \ \tan \left( {-\displaystyle \frac{{\theta -\phi }}{2}} \right)\ \ \ \left[ {\because \tan \left( {-\theta } \right)=-\tan \theta } \right]\\\\\ \ \ \ =\ \ \tan \displaystyle \frac{{\phi -\theta }}{2}\ \end{array}$


$ \displaystyle \begin{array}{l}\text{(j)}\ \ \ \ \displaystyle \frac{{\sin \alpha +\sin \beta }}{{\cos \alpha -\cos \beta }}\\\\\ \ \ \ =\ \ \displaystyle \frac{{2\sin \displaystyle \frac{{\alpha +\beta }}{2}\cos \displaystyle \frac{{\alpha -\beta }}{2}}}{{-2\sin \displaystyle \frac{{\alpha +\beta }}{2}\sin \displaystyle \frac{{\alpha -\beta }}{2}}}\\\\\ \ \ \ =\ \ \displaystyle \frac{{\cos \displaystyle \frac{{\alpha -\beta }}{2}}}{{-\sin \displaystyle \frac{{\alpha -\beta }}{2}}}\\\\\ \ \ \ =\ \ -\cot \displaystyle \frac{{\alpha -\beta }}{2}\\\\\ \ \ \ =\ \ \cot \left( {-\displaystyle \frac{{\alpha -\beta }}{2}} \right)\ \ \ \left[ {\because \cot \left( {-\theta } \right)=-\cot \theta } \right]\\\\\ \ \ \ =\ \ \cot \displaystyle \frac{{\beta -\alpha }}{2}\ \end{array}$


$ \displaystyle \begin{array}{l}\text{(k)}\ \ \ \ \displaystyle \frac{{\sin 2x}}{{1+\cos 2x}}\\\\\ \ \ \ =\ \ \displaystyle \frac{{2\sin x\cos x}}{{1+2{{{\cos }}^{2}}x-1}}\\\\\ \ \ \ =\ \ \displaystyle \frac{{2\sin x\cos x}}{{2{{{\cos }}^{2}}x}}\\\\\ \ \ \ =\ \ \displaystyle \frac{{\sin x}}{{\cos x}}\\\\\ \ \ \ =\ \ \tan x\end{array}$


$ \displaystyle \begin{array}{l}\text{(l)}\ \ \ \ \displaystyle \frac{{\sin 3\alpha }}{{\cos \alpha }}+\displaystyle \frac{{\cos 3\alpha }}{{\sin \alpha }}\\\\\ \ \ \ =\ \ \displaystyle \frac{{\sin 3\alpha \sin \alpha +\cos 3\alpha \cos \alpha }}{{\sin \alpha \cos \alpha }}\\\\\ \ \ \ =\ \ \displaystyle \frac{{\cos 3\alpha \cos \alpha +\sin 3\alpha \sin \alpha }}{{\sin \alpha \cos \alpha }}\\\\\ \ \ \ =\ \ \displaystyle \frac{{2\cos \left( {3\alpha -\alpha } \right)}}{{2\sin \alpha \cos \alpha }}\\\\\ \ \ \ =\ \ \displaystyle \frac{{2\cos 2\alpha }}{{\sin 2\alpha }}\\\\\ \ \ \ =\ \ 2\cot 2\alpha \end{array}$


$ \displaystyle \begin{array}{l}\text{(m)}\ \ \ \ \displaystyle \frac{1}{{\cos 2\alpha }}-\displaystyle \frac{{\cos 2\alpha }}{{1+\sin 2\alpha }}\\\\\ \ \ \ =\ \ \displaystyle \frac{{1+\sin 2\alpha -{{{\cos }}^{2}}2\alpha }}{{\cos 2\alpha \left( {1+\sin 2\alpha } \right)}}\\\\\ \ \ \ =\ \ \displaystyle \frac{{\sin 2\alpha +1-{{{\cos }}^{2}}2\alpha }}{{\cos 2\alpha \left( {1+\sin 2\alpha } \right)}}\\\\\ \ \ \ =\ \ \displaystyle \frac{{\sin 2\alpha +{{{\sin }}^{2}}2\alpha }}{{\cos 2\alpha \left( {1+\sin 2\alpha } \right)}}\\\\\ \ \ \ =\ \ \displaystyle \frac{{\sin 2\alpha \left( {1+\sin 2\alpha } \right)}}{{\cos 2\alpha \left( {1+\sin 2\alpha } \right)}}\\\\\ \ \ \ =\ \ \tan 2\alpha \end{array}$


$ \displaystyle \begin{array}{l}\text{(n)}\ \ \ \ \displaystyle \frac{{\tan \alpha }}{{\sec \alpha -1}}+\displaystyle \frac{{\tan \alpha }}{{\sec \alpha +1}}\\\\\ \ \ \ =\ \ \displaystyle \frac{{\tan \alpha \left( {\sec \alpha +1} \right)+\tan \alpha \left( {\sec \alpha -1} \right)}}{{{{{\sec }}^{2}}\alpha -1}}\\\\\ \ \ \ =\ \ \displaystyle \frac{{\tan \alpha \left( {\sec \alpha +1+\sec \alpha -1} \right)}}{{{{{\tan }}^{2}}\alpha }}\\\ \ \ \ \ \ \ \ \ \ \ \ \left[ {\because {{{\tan }}^{2}}\alpha +1={{{\sec }}^{2}}\alpha } \right]\\\\\ \ \ \ =\ \ \displaystyle \frac{{2\tan \alpha \sec \alpha }}{{{{{\tan }}^{2}}\alpha }}\\\\\ \ \ \ =\ \ \displaystyle \frac{{2\sec \alpha }}{{\tan \alpha }}\\\\\ \ \ \ =\ \ \displaystyle \frac{2}{{\cos \alpha }}\times \displaystyle \frac{{\cos \alpha }}{{\sin \alpha }}\\\\\ \ \ \ =\ \ \displaystyle \frac{2}{{\sin \alpha }}\\\\\ \ \ \ =\ \ 2\operatorname{cosec}\alpha \end{array}$


$ \displaystyle \begin{array}{l}\text{(o)}\ \ \ \ \displaystyle \frac{{2-{{{\operatorname{cosec}}}^{2}}\alpha }}{{{{{\operatorname{cosec}}}^{2}}\alpha +2\cot \alpha }}\\\\\ \ \ \ =\ \ \displaystyle \frac{{2-\displaystyle \frac{1}{{{{{\sin }}^{2}}\alpha }}}}{{\displaystyle \frac{1}{{{{{\sin }}^{2}}\alpha }}+2\left( {\displaystyle \frac{{\cos \alpha }}{{\sin \alpha }}} \right)}}\\\\\ \ \ \ =\ \ \displaystyle \frac{{\displaystyle \frac{{2{{{\sin }}^{2}}\alpha -1}}{{{{{\sin }}^{2}}\alpha }}}}{{\displaystyle \frac{{1+2\sin \alpha \cos \alpha }}{{{{{\sin }}^{2}}\alpha }}}}\\\\\ \ \ \ =\ \ \displaystyle \frac{{2{{{\sin }}^{2}}\alpha -1}}{{1+2\sin \alpha \cos \alpha }}\\\\\ \ \ \ =\ \ \displaystyle \frac{{2{{{\sin }}^{2}}\alpha -{{{\sin }}^{2}}\alpha -{{{\cos }}^{2}}\alpha }}{{{{{\sin }}^{2}}\alpha +2\sin \alpha \cos \alpha +{{{\cos }}^{2}}\alpha }}\\\ \ \ \ \ \ \ \ \ \ \left[ {\because {{{\sin }}^{2}}\alpha +{{{\cos }}^{2}}\alpha =1} \right]\\\\\ \ \ \ =\ \ \displaystyle \frac{{{{{\sin }}^{2}}\alpha -{{{\cos }}^{2}}\alpha }}{{{{{\left( {\sin \alpha +\cos \alpha } \right)}}^{2}}}}\\\\\ \ \ \ =\ \ \displaystyle \frac{{\left( {\sin \alpha -\cos \alpha } \right)\left( {\sin \alpha +\cos \alpha } \right)}}{{{{{\left( {\sin \alpha +\cos \alpha } \right)}}^{2}}}}\\\\\ \ \ \ =\ \ \displaystyle \frac{{\sin \alpha -\cos \alpha }}{{\sin \alpha +\cos \alpha }}\end{array}$


$ \displaystyle \begin{array}{l}\text{(p)}\ \ \ \ 1-\displaystyle \frac{{{{{\sin }}^{2}}x}}{{1+\cos x}}\\\\\ \ \ \ =\ \ \displaystyle \frac{{1+\cos x-{{{\sin }}^{2}}x}}{{1+\cos x}}\\\\\ \ \ \ =\ \ \displaystyle \frac{{\cos x+1-{{{\sin }}^{2}}x}}{{1+\cos x}}\\\\\ \ \ \ =\ \ \displaystyle \frac{{\cos x+{{{\cos }}^{2}}x}}{{1+\cos x}}\\\\\ \ \ \ =\ \ \displaystyle \frac{{\cos x\left( {1+\cos x} \right)}}{{1+\cos x}}\\\\\ \ \ \ =\ \ \cos x\end{array}$


$ \displaystyle \begin{array}{l}\text{(q)}\ \ \ \ \displaystyle \frac{{1+{{{\cot }}^{2}}x}}{{1-{{{\cot }}^{2}}x}}\\\\\ \ \ \ =\ \ \displaystyle \frac{{1+\displaystyle \frac{{{{{\cos }}^{2}}x}}{{{{{\sin }}^{2}}x}}}}{{1-\displaystyle \frac{{{{{\cos }}^{2}}x}}{{{{{\sin }}^{2}}x}}}}\\\\\ \ \ \ =\ \ \displaystyle \frac{{\displaystyle \frac{{{{{\sin }}^{2}}x+{{{\cos }}^{2}}x}}{{{{{\sin }}^{2}}x}}}}{{\displaystyle \frac{{{{{\sin }}^{2}}x-{{{\cos }}^{2}}x}}{{{{{\sin }}^{2}}x}}}}\\\\\ \ \ \ =\ \ \displaystyle \frac{{{{{\sin }}^{2}}x+{{{\cos }}^{2}}x}}{{{{{\sin }}^{2}}x-{{{\cos }}^{2}}x}}\\\\\ \ \ \ =\ \ -\displaystyle \frac{1}{{{{{\cos }}^{2}}x-{{{\sin }}^{2}}x}}\\\\\ \ \ \ =\ \ -\displaystyle \frac{1}{{\cos 2x}}\\\\\ \ \ \ =\ \ -\sec 2x\end{array}$


$ \displaystyle \begin{array}{l}\text{(r)}\ \ \ \ \displaystyle \frac{{\cos \alpha +\sec \beta }}{{\cos \beta +\sec \alpha }}\\\\\ \ \ \ =\ \ \displaystyle \frac{{\cos \alpha +\displaystyle \frac{1}{{\cos \beta }}}}{{\cos \beta +\displaystyle \frac{1}{{\cos \alpha }}}}\\\\\ \ \ \ =\ \ \displaystyle \frac{{\displaystyle \frac{{\cos \alpha \cos \beta +1}}{{\cos \beta }}}}{{\displaystyle \frac{{\cos \alpha \cos \beta +1}}{{\cos \alpha }}}}\\\\\ \ \ \ =\ \ \displaystyle \frac{1}{{\cos \beta }}\times \cos \alpha \\\\\ \ \ \ =\ \ \cos \alpha \sec \beta \end{array}$