If $\sin^{-1} (x) + \sin^{-1} (y)+ \sin^{-1} (z)=\dfrac {\pi}{2}$, prove that $x^2+y^2+z^2+2xyz=1$

Let: $\sin^{-1}x=\alpha$, $\sin^{-1}y=\beta$, $\sin^{-1}z=\gamma$. We know that $\alpha+\beta+\gamma={\pi\over2}$, that is: $\gamma={\pi\over2}-\alpha-\beta$ and $$\sin\gamma=\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta.$$ We have then: $$ \begin{align} &x^2+y^2+z^2+2xyz=\\ &\sin^2\alpha+\sin^2\beta+\sin^2\gamma+2\sin\alpha\sin\beta\sin\gamma=\\ &\sin^2\alpha+\sin^2\beta+(\cos\alpha\cos\beta-\sin\alpha\sin\beta)^2 +2\sin\alpha\sin\beta(\cos\alpha\cos\beta-\sin\alpha\sin\beta)=\\ &\sin^2\alpha+\sin^2\beta+\cos^2\alpha\cos^2\beta-\sin^2\alpha\sin^2\beta=\\ &\sin^2\alpha(1-\sin^2\beta)+\sin^2\beta+\cos^2\alpha\cos^2\beta=\\ &\sin^2\alpha\cos^2\beta+\sin^2\beta+\cos^2\alpha\cos^2\beta=\\ &(\sin^2\alpha+\cos^2\alpha)\cos^2\beta+\sin^2\beta=\\ &\cos^2\beta+\sin^2\beta=1 \end{align} $$