Answer
$tan(sin^{-1}x)
= \frac{x}{\sqrt {1-x^{2}}}$
Work Step by Step
$tan(sin^{-1}x)$ = $\frac {sin(sin^{-1}(x))}{cos(sin^{-1}(x))}$
$sin(sin^{-1}(x))$=$x$
$cos(sin^{-1}(x))$=$\sqrt{1-sin^2(sin^{-1}(x))}$=$\sqrt{1-x^2}$
$tan(sin^{-1}x)$ = $\frac{x}{\sqrt {1-x^{2}}}$
