A) \[2\sin \alpha \]
B) \[2\sin \alpha cos\beta sin\gamma \]
C) \[2\sin \alpha \sin \beta \cos \gamma \]
D) \[2\sin \alpha \sin \beta \sin \gamma \]
Correct Answer: C
Solution :
\[\therefore \]\[{{\sin }^{2}}\alpha +\sin (\beta -\gamma )\sin (\beta +\gamma )\] \[={{\sin }^{2}}\alpha +\sin (\pi -\alpha )\sin (\beta -\gamma )\] \[=\sin \alpha [\sin \alpha +\sin (\beta -\gamma )]\] \[=\sin \alpha [\sin (\pi -(\beta +\gamma ))+\sin (\beta -\gamma )]\] \[=\sin \alpha [\sin (\beta -\gamma )+\sin (\beta +\gamma )]\] \[=2\sin \alpha \sin \beta \cos \gamma \]
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