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If $\alpha+\beta+\gamma=\pi$ then the value of $\sin^2\alpha+\sin^2\beta-\sin^2\gamma$ is equal to

$\begin{array}{1 1}(a)\;2\sin\alpha &(b)\;2\sin\alpha\cos\beta\sin\gamma\\(c)\;2\sin\alpha\sin\beta\cos\gamma&(d)\;2\sin\alpha\sin\beta\sin\gamma\end{array}$

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1 Answer

$\sin^2\alpha+\sin(\beta-\gamma)\sin(\beta+\gamma)=\sin^2\alpha+\sin(\pi-\alpha)\sin(\beta-\gamma)$
$\qquad\qquad\qquad\qquad\qquad\quad=\sin[\sin\alpha+\sin(\beta-\gamma)+\sin(\beta+\gamma)]$
$\qquad\qquad\qquad\qquad\qquad\quad=2\sin\alpha\sin\beta\cos\gamma$
Hence (c) is the correct answer.
answered Oct 7, 2013 by sreemathi.v
 

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