Answer
$\frac{\sqrt {1-u^2}}{u}$
Work Step by Step
Step 1. Letting $sec^{-1}u=t$, we have $sec(t)=u$ and $0\le t \le \pi$
Step 2. Form a right triangle with sides $x=|u|,r=1,y=\sqrt {1-u^2}$.
Step 3. As $tan(t)$ and $sec(t)$ have the same sign in the above interval, we have $tan(sec^{-1}u)=tan(t)=\frac{\sqrt {1-u^2}}{u}$