Chapter 7 - Analytic Trigonometry - 7.2 The Inverse Trigonometric Functions (Continued) - 7.2 Assess Your Understanding - Page 458: 27

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

Step 1. Letting $tan^{-1}\frac{1}{2}=t$, we have $tan(t)=\frac{1}{2}$ 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 have $sec(t)=\frac{r}{x}=\frac{\sqrt 5}{2}$ or $sec(tan^{-1}\frac{1}{2})=\frac{\sqrt 5}{2}$