<s>[INST] If $\sin^{-1} (x) + \sin^{-1} (y)+ \sin^{-1} (z)=\dfrac {\pi}{2}$, prove that $x^2+y^2+z^2+2xyz=1$ 
If $\sin^{-1} (x) + \sin^{-1} (y)+ \sin^{-1} (z)=\dfrac {\pi}{2}$, prove that:
  $$x^2+y^2+z^2+2xyz=1$$

My Attempt:
$$\sin^{-1} (x) + \sin^{-1} (y) + \sin^{-1} (z)=\dfrac {\pi}{2}$$
$$\sin^{-1} (x\sqrt {1-y^2}+y\sqrt {1-x^2})+\sin^{-1} (z)=\dfrac {\pi}{2}$$
$$\sin^{-1} (x\sqrt {1-y^2} + y\sqrt {1-x^2})=\dfrac {\pi}{2} - \sin^{-1} (z)$$
 [/INST] Put differently, if $\alpha+\beta+\gamma=\frac \pi2$, then 
$$ \tag1\sin^2\alpha+\sin^2\beta+\sin^2\gamma+2\sin\alpha\sin\beta\sin\gamma=1.$$
Let's investigate how the left hand side aries if we keep $\gamma $ fixed, that is, we consider 
$$\tag2 \sin^2(\alpha+t)+\sin^2(\beta-t)+\sin^2\gamma+2\sin(\alpha+t)\sin(\beta-t)\sin\gamma$$
and compute its derivative $\frac d{dt}$ at $t=0$:
$$ \begin{align}&2\sin \alpha\cos \alpha-2\sin \beta\cos\beta+2(\cos\alpha\sin\beta-\sin\alpha\cos\beta)\sin\gamma\\
={}&\sin2\alpha-\sin 2\beta-2\sin(\alpha-\beta)\sin\gamma\\
{}={}&2\cos(\alpha+\beta)\sin(\alpha-\beta)-2\sin(\alpha-\beta)\sin\gamma\\
={}&2\sin\gamma\sin(\alpha-\beta)-2\sin(\alpha-\beta)\sin\gamma\\={}&0\end{align}$$
We conclude that $(2)$ is constant as a function of $t$. Therefore, it suffices to show $(1)$ for the case $\beta=0$ (i.e., $t=\beta$), i.e., we are reduced to showing

If $\alpha+\gamma=\frac\pi2$ then $\sin^2\alpha+\sin^2\gamma=1$.

But that is of course true.
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