Chapter 6 - Analytic Trigonometry - Section 6.2 The Inverse Trigonometric Functions (Continued) - 6.2 Assess Your Understanding - Page 480: 34

$\sqrt 5$.

1. Let $tan^{-1}(\frac{1}{2})=t$, we have $tan(t)=\frac{1}{2}$ and $t$ in quadrant I. 2. Let $x=2, y=1$, we have $r=\sqrt {(2)^2+(1)^2}=\sqrt {5}$. 3. Thus $csc(tan^{-1}(\frac{1}{2}))=csc(t)=\frac{r}{y}=\sqrt 5$.