Chapter 7 - Analytic Trigonometry - 7.2 The Inverse Trigonometric Functions (Continued) - 7.2 Assess Your Understanding - Page 458: 25

$\frac{\sqrt 2}{4}$

Step 1. Letting $sin^{-1}\frac{1}{3}=t$, we have $sin(t)=\frac{1}{3}$ Step 2. $y=1, r=3$, and $x=\sqrt {r^2-y^2}=2\sqrt 2$. Step 3. Sides $x,y,r$ form a right triangle with angle $t$ facing $y$. Step 4. We have $tan(t)=\frac{y}{x}=\frac{\sqrt 2}{4}$ or $tan(sin^{-1}\frac{1}{3})=\frac{\sqrt 2}{4}$