Answer
$ \sqrt 5$
Work Step by Step
Step 1. Letting $tan^{-1}(\frac{1}{2})=t$, we have $tan(t)=\frac{1}{2}$ and $0\lt t\lt\frac{\pi}{2}$.
Step 2. Let $x=2, y=1$, and $r=\sqrt {x^2+y^2}=\sqrt {5}$.
Step 3. Sides $x,y,r$ form a right triangle with angle $|t|$ facing $y$.
Step 4. We know $csc(t)$ is positive and we have $csc(t)=\frac{r}{y}=\sqrt 5$ or $csc(tan^{-1}(\frac{1}{2}))= \sqrt 5$