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यदि $\alpha + \beta - \gamma = \pi ,$ तो ${\sin ^2}\alpha + {\sin ^2}\beta - {\sin ^2}\gamma $ बराबर है
$2\,\sin \alpha \,\sin \beta \,\cos \gamma $
$ 2\,\cos \alpha \,\cos \beta \,\cos \gamma$
$2\,\sin \alpha \,\sin \beta \sin \gamma $
इनमें से कोई नहीं
Solution
(a) यहाँ $\alpha + \beta – \gamma = \pi .$
अब ${\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 $.
