2016/047) if $\alpha+\beta+\gamma = \pi$ then prove that $\sin ^2\alpha + \sin ^2 \beta - \sin ^2 \gamma = 2 \sin\,\alpha\sin\,\beta \cos\,\gamma$

$\sin ^2\alpha + \sin ^2 \beta - \sin ^2 \gamma$
$= \sin ^2\alpha + \sin^2  \beta - \sin ^2 \gamma(\sin ^2 \beta + \cos  ^2 \beta)$
$= \sin ^2\alpha + \sin^2  \beta(1 - \sin ^2 \gamma) - \sin ^2 \gamma \cos  ^2 \beta$
$= \sin ^2\alpha + \sin^2  \beta\cos ^2 \gamma - \sin ^2 \gamma \cos  ^2 \beta$
$= \sin ^2\alpha + (\sin\, \beta\cos\, \gamma + \sin\, \gamma \cos\, \beta)(\sin\, \beta\cos\, \gamma - \sin\, \gamma \cos\, \beta)$
$= \sin ^2\alpha + (\sin(\beta + \gamma)\sin(\beta- \gamma)$
$=\sin\,\alpha \sin (\sin(\beta + \gamma) + \sin\,\alpha\sin(\beta - \gamma)$ as $\sin\,\alpha= \sin(\beta + \gamma)$
$=\sin\,\alpha (\sin(\beta + \gamma)+\sin(\beta - \gamma))$
$=\sin\,\alpha (2\sin\,\beta \sin\,\gamma)$
$=2 \sin\,\alpha \sin\,\beta \sin\,\gamma$