(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 $.