<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] I want to show a slight variation of Aretino's nice proof that does not presuppose knowledge of the identity we want to prove, but rather derives it from the given equation. We need the identity $\cos(\sin^{-1}(a)) = \sqrt{1-a^2}$ as well as
$$ \sin(a+b+c) =  \sum\limits_{cyc} \Big(\sin(a)\cos(b)\cos(c)\Big)-\sin(a)\sin(b)\sin(c) $$
Thus by applying $\sin$ on both sides of $\frac \pi2 = \sin^{-1} (x) + \sin^{-1} (y)+ \sin^{-1} (z)$ we deduce
$$ 1 = \sum_{cyc} \Big( x\sqrt{1-y^2}\sqrt{1-z^2}\Big)- xyz  $$
On the other hand if we solve the original equation for any of the variables, say, $x$ we obtain:
$$ x = \sqrt{1-y^2}\sqrt{1-z^2} - yz $$
Which we can plug into the former equation to eliminate the roots:
$$ 1 = \sum_{cyc} \Big( x(x+yz)\Big)- xyz = x^2+y^2+z^2 +2xyz$$
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