Answer
$\frac{1}{\sqrt {1+u^2}}=\frac{\sqrt {1+u^2}}{1+u^2}$
Work Step by Step
Step 1. Letting $tan^{-1}u=t$, we have $tan(t)=u$
Step 2. Form a right triangle with sides $x=1,y=u,r=\sqrt {1+u^2}$ and angle $|t|$ facing $y$.
Step 3. We have $cos(tan^{-1}u)=cos(t)=\frac{x}{r}=\frac{1}{\sqrt {1+u^2}}=\frac{\sqrt {1+u^2}}{1+u^2}$