Answer
$\frac{u\sqrt {1-u^2}}{1-u^2}$
Work Step by Step
Let $v=arcsin(u)$, we have $sin(v)=u$. Form a right triangle of sides $u, \sqrt {1-u^2},1$ with angle $v$ facing the side $u$, we have $tan(v)=\frac{u}{\sqrt {1-u^2}}$. Thus $tan(arcsin(u))=tan(v)=\frac{u}{\sqrt {1-u^2}}=\frac{u\sqrt {1-u^2}}{1-u^2}$
