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$.
Show/Hide Solution
| $ \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 )$.
Show/Hide Solution
| $ \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)$.
Show/Hide Solution
| $ \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}}}}$
(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}}}}$
Show/Hide Solution
| $ \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$
(a) $ \displaystyle \text{sin 47}{}^\circ +\cos 77{}^\circ =\cos 17{}^\circ $
(b) $ \displaystyle \cos 80{}^\circ +\cos 40{}^\circ -\cos 20{}^\circ =0$
Show/Hide Solution
| $ \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)$.
Show/Hide Solution
| $ \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)$.
Show/Hide Solution
| $ \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 $
Show/Hide Solution
| $ \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)$.
Show/Hide Solution
| $ \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 $
(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 $
Show/Hide Solution
| $ \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$.
Show/Hide Solution
| $ \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$.
Show/Hide Solution
| $ \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 $.
Show/Hide Solution
| $ \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$.
Show/Hide Solution
| $ \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$.
Show/Hide Solution
| $ \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 }}$.
Show/Hide Solution
| $ \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}}$.
Show/Hide Solution
| $ \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 $.
Show/Hide Solution
| $ \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)$.
Show/Hide Solution
| $ \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.
Show/Hide Solution
| $ \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}$ |
0 Reviews:
Post a Comment