Chapter 7 - Analytic Trigonometry - 7.2 The Inverse Trigonometric Functions (Continued) - 7.2 Assess Your Understanding - Page 458: 33

$ \sqrt 5$

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