MCQ
If $\alpha + \beta - \gamma = \pi ,$ then ${\sin ^2}\alpha + {\sin ^2}\beta - {\sin ^2}\gamma = $
  • $2\,\sin \alpha \,\sin \beta \,\cos \gamma $
  • B
    $ 2\,\cos \alpha \,\cos \beta \,\cos \gamma$
  • C
    $2\,\sin \alpha \,\sin \beta \sin \gamma $
  • D
    None of these

Answer

Correct option: A.
$2\,\sin \alpha \,\sin \beta \,\cos \gamma $
a
(a) We have $\alpha + \beta - \gamma = \pi .$ 

Now ${\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 )$  

                                                            $(\because \alpha  + \beta  - \gamma  = \pi )$ 

$ = {\sin ^2}\alpha + \sin \alpha \sin (\beta + \gamma ) = \sin \alpha \{ \sin \alpha + \sin (\beta + \gamma )\} $ 

$ = \sin \alpha \{ \sin (\pi - \overline {\beta + \gamma )} + \sin (\beta + \gamma )\} $

$ = \sin \alpha \{ - \sin (\gamma - \beta ) + \sin (\gamma + \beta )\} $

$ = \sin \alpha \{ 2\sin \beta \cos \gamma \} = 2\sin \alpha \sin \beta \cos \gamma $.

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