Chapter 6 - Analytic Trigonometry - Section 6.2 The Inverse Trigonometric Functions (Continued) - 6.2 Assess Your Understanding - Page 480: 33

$\sqrt 5$.

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