Answer
$cos(tan^{-1}~2)=\frac{\sqrt 5}{5}$
Work Step by Step
Let $u=tan^{-1}~2$. Then:
$tan~u=2$
The range $tan^{-1}x$ is $-\frac{\pi}{2}\lt x\lt\frac{\pi}{2}$. So, since $tan^{-1}~2\gt0$, then $0\lt u\lt\frac{\pi}{2}$ (First Quadrant)
$sec^2u=1+tan^2u$
$sec^2u=1+2^2=5$
$sec~u=\sqrt 5~~$ (First Quadrant)
$cos(tan^{-1}~2)=cos~u=\frac{1}{sec~u}=\frac{1}{\sqrt 5}=\frac{1}{\sqrt 5}\frac{\sqrt 5}{\sqrt 5}=\frac{\sqrt 5}{5}$
