Answer
$\displaystyle \frac{x}{\sqrt{1-x^{2}}}$
Work Step by Step
Let $t=\sin^{-1}x.$
Then, $t\in[- \displaystyle \frac{\pi}{2},\frac{\pi}{2}]$ and $\sin t=x$
$\displaystyle \tan t=\frac{\sin t}{\cos t}=\frac{x}{\cos(\sin^{-1}x)}$
(the numerator is x, because of the way we defined $t$.)
The denominator equals $\sqrt{1-x^{2}}$ as found in the previous exercise.
$\displaystyle \tan t=\frac{x}{\sqrt{1-x^{2}}}$
