$\sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z=\frac{\pi}{2}$
Put $\sin ^{-1} x=\alpha, \quad \sin ^{-1} y=\beta, \quad \sin ^{-1} z=\gamma$
$\begin{aligned}
& \therefore \alpha+\beta+\gamma=\frac{\pi}{2} \text {, } \\
& \alpha+\beta=\frac{\pi}{2}-\gamma \quad \text { or } \cos (\alpha+\beta)=\cos \left(\frac{\pi}{2}-\gamma\right) \\
&
\end{aligned}$
So,
$\cos \alpha \cos \beta-\sin \alpha \sin \beta=\sin \gamma$
and, we have $\sin \alpha=x \Rightarrow \cos \alpha=\sqrt{1-x^2}$
Similarly, $\quad \cos \beta=\sqrt{1-y^2}$
$\therefore$ From equation (i), we get $\sqrt{1-x^2} \cdot \sqrt{1-y^2}=x y+z$
Squaring both sides, we have $x^2+y^2+z^2+2 x y z=1$.