If $\alpha+\beta+\gamma=\pi$, then the expression $\sin ^2 \alpha+\sin ^2 \beta-\sin ^2 \gamma$ has the value
If $\alpha+\beta+\gamma=\pi$, then the expression $\sin ^2 \alpha+\sin ^2 \beta-\sin ^2 \gamma$ has the value
- $2 \sin \alpha \sin \beta \sin \gamma$
- $2 \cos \alpha \sin \beta \sin \gamma$
- $2 \sin \alpha \cos \beta \sin \gamma$
- $2 \sin \alpha \sin \beta \cos \gamma$
Solution
$\begin{aligned} & \sin ^2 \alpha+\sin ^2 \beta-\sin ^2 \gamma \\ & =\sin ^2 \alpha+\sin (\beta+\gamma) \sin (\beta-\gamma) \\ & =\sin ^2 \alpha+\sin (\pi-\alpha) \sin (\beta-\gamma) \\ & =\sin \alpha[\sin \alpha+\sin (\beta-\gamma)] \\ & =\sin \alpha[\sin (\beta+\gamma)+\sin (\beta-\gamma)] \\ & =2 \sin \alpha \sin \beta \cos \gamma\end{aligned}$
Asked in: MHT CET 2024 (02 May Shift 2)
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