Chapter 6 - Analytic Trigonometry - Section 6.2 The Inverse Trigonometric Functions (Continued) - 6.2 Assess Your Understanding - Page 480: 27

$\frac{\sqrt 5}{2}$.

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 $sec(tan^{-1}\frac{1}{2})=sec(t)=\frac{r}{x}=\frac{\sqrt 5}{2}$.