Exercise (11.3) No(4, 5) - Solution

4. If $ \displaystyle α + β + γ = 180°$, prove that

$ \displaystyle \text{(a)}\ \ \sin (\alpha +\beta )=\cos (90{}^\circ -\gamma )$

$ \displaystyle \text{(b)}\ \ \sin (\frac{{\alpha +\beta }}{2})=\sin (90{}^\circ +\frac{\gamma }{2})$

$ \displaystyle \text{(c)}\ \ \tan \left( {\frac{\alpha }{2}} \right)=\cot \left( {180{}^\circ +\frac{{\beta +\gamma }}{2}} \right)$

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$ \displaystyle \text{(a)}\ \ \alpha +\beta +\gamma =180{}^\circ $

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \alpha +\beta =180{}^\circ -\gamma \\\\\therefore \ \ \ \ \sin (\alpha +\beta )=\sin (180{}^\circ -\gamma )\\\ \ \ \\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\sin \gamma \\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ \cos (90{}^\circ -\gamma )\end{array}$


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


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

5.    Prove that in any triangle $ \displaystyle ABC,$

(i) $ \displaystyle \sin (A+B) = \sin C.$

(ii) $ \displaystyle \cos(A+B) + \cos C = 0.$

(iii) $ \displaystyle \cos \frac{A+B}{2} = \sin \frac{C}{2}.$

(iv) $ \displaystyle \tan \frac{A+B}{2} = \cot \frac{C}{2}.$

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(i) $ \displaystyle \text{Since}\ A+B+C=180{}^\circ ,$

$ \displaystyle \begin{array}{l}\therefore A+B=180{}^\circ -C\\\\\therefore \sin (A+B)=\sin (180{}^\circ -C)\\\\\therefore \sin (A+B)=\sin C\end{array}$


(ii) $ \displaystyle \text{Similarly, }\cos (A+B)=\cos (180{}^\circ -C)$

$ \displaystyle \begin{array}{l}\therefore \cos (A+B)=-\cos C\\\\\therefore \cos (A+B)+\cos C=0\end{array}$


(iii) $ \displaystyle \cos \left( {\frac{{A+B}}{2}} \right)=\cos \left( {\frac{{180{}^\circ -C}}{2}} \right)$

$ \displaystyle \therefore \ \cos \left( {\frac{{A+B}}{2}} \right)=\cos \left( {90{}^\circ -\frac{C}{2}} \right)$

$ \displaystyle \therefore \ \cos \left( {\frac{{A+B}}{2}} \right)=\sin \frac{C}{2}$


(iv) $ \displaystyle \tan \left( {\frac{{A+B}}{2}} \right)=\tan \left( {\frac{{180{}^\circ -C}}{2}} \right)$

$ \displaystyle \therefore \ \tan \left( {\frac{{A+B}}{2}} \right)=\tan \left( {90{}^\circ -\frac{C}{2}} \right)$

$ \displaystyle \therefore \ \tan \left( {\frac{{A+B}}{2}} \right)=\cot \frac{C}{2}$