Trigonometrical Ratios and Identities 1 Question 28
28. Prove that $\sin ^{2} \alpha+\sin ^{2} \beta-\sin ^{2} \gamma=2 \sin \alpha \sin \beta \sin \gamma$, where $\alpha+\beta+\gamma=\pi$.
$(1978,4 M)$
Integer Answer Type Question
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Solution:
- LHS $=\sin ^{2} \alpha+\sin ^{2} \beta-\sin ^{2} \gamma$
$$ \begin{aligned} & =\sin ^{2} \alpha+\left(\sin ^{2} \beta-\sin ^{2} \gamma\right) \\ & =\sin ^{2} \alpha+\sin (\beta+\gamma) \sin (\beta-\gamma) \\ & =\sin ^{2} \alpha+\sin (\pi-\alpha) \sin (\beta-\gamma) \quad[\because \alpha+\beta+\gamma=\pi] \\ & =\sin ^{2} \alpha+\sin \alpha \sin (\beta-\gamma) \\ & =\sin \alpha[\sin \alpha+\sin (\beta-\gamma)] \\ & =\sin \alpha[\sin (\pi-(\beta+\gamma))+\sin (\beta-\gamma)] \\ & =\sin \alpha[\sin (\beta+\gamma)+\sin (\beta-\gamma)] \\ & =\sin \alpha[2 \sin \beta \cos \gamma] \\ & =2 \sin \alpha \sin \beta \cos \gamma=\text { RHS } \end{aligned} $$
