Answer
The identity is verified.
$tan(sin^{-1}x)=\frac{x}{\sqrt {1-x^2}}$
Work Step by Step
$y=sin^{-1}x$, domain: $-1\leq x\leq1$, range: $-\frac{\pi}{2}\leq y\leq\frac{\pi}{2}$:
$x=sin~y$
$x^2=sin^2y$
$1-x^2=1-sin^2y$
$1-x^2=cos^2y$
$cos~y=+\sqrt {1-x^2}~~$ (Use only the $+$ because $-\frac{\pi}{2}\leq y\leq\frac{\pi}{2}$)
$tan(sin^{-1}x)=\frac{sin(sin^{-1}x)}{cos(sin^{-1}x)}=\frac{sin~y}{cos~y}=\frac{x}{\sqrt {1-x^2}}$
