triangle trigonometry In triangle $ABC$,if $\sin^2 \alpha+ \sin^2 \beta = \sin \gamma$ and  $\alpha$ and $\beta$ are acute show that $\gamma= 90 $ degree. $$ $$
I try following:
$$\dfrac{1-\cos 2\alpha}{2}+\dfrac{1-\cos 2 \beta}{2}=\sin (\alpha+\beta) $$ 
$$1-\cos(\alpha+\beta)\cos(\alpha-\beta)=\sin(\alpha+\beta)$$
Now,i want to factorize one side of equation,and on the other i want to have 0 or 1.But i don't know how to do it
 A: $$1-\sin\gamma=1-[\sin^2\alpha+\sin^2\beta]=\cos^2\alpha-\sin^2\beta=\cos(\alpha+\beta)\cos(\alpha-\beta)$$
Now $\cos(\alpha+\beta)=\cos(\pi-\gamma)=-\cos\gamma$
$$\implies1-\sin\gamma=-\cos\gamma\cos(\alpha-\beta)$$
$$\iff\left(\cos\dfrac\gamma2-\sin\dfrac\gamma2\right)^2+\left(\cos\dfrac\gamma2-\sin\dfrac\gamma2\right)\left(\cos\dfrac\gamma2+\sin\dfrac\gamma2\right)\cos(\alpha-\beta)=0$$
$$\iff\left(\cos\dfrac\gamma2-\sin\dfrac\gamma2\right)\left[\left(\cos\dfrac\gamma2-\sin\dfrac\gamma2\right)+\left(\cos\dfrac\gamma2+\sin\dfrac\gamma2\right)\cos(\alpha-\beta)\right]=0$$
If $\cos\dfrac\gamma2-\sin\dfrac\gamma2=0\iff\tan\dfrac\gamma2=1=\tan\dfrac\pi4\iff\dfrac\gamma2=n\pi+\dfrac\pi4\iff\gamma=2n\pi+\dfrac\pi2$ where $n$ is any integer
As $0<\gamma<\pi,n=0$
Else $\left(\cos\dfrac\gamma2-\sin\dfrac\gamma2\right)+\left(\cos\dfrac\gamma2+\sin\dfrac\gamma2\right)\cos(\alpha-\beta)=0$
$\iff\cos(\alpha-\beta)=\cdots=\tan\left(\dfrac\gamma2-\dfrac\pi4\right)\  \ \ \ (1)$
We need $-1<\tan\left(\dfrac\gamma2-\dfrac\pi4\right)\le1$
Clearly, $(1)$ does not have any unique solution.
