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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