Chapter 7 - Analytic Trigonometry - 7.2 The Inverse Trigonometric Functions (Continued) - 7.2 Assess Your Understanding - Page 458: 74

$\frac{5}{13}$

Step 1. Letting $h^{-1}(\frac{5}{12})=tan^{-1}(\frac{5}{12})=t$, we have $tan(t)=\frac{5}{12}$ and $t$ in quadrant I. Step 2. Form a right triangle with sides $x=12, y=5, r=\sqrt {x^2+y^2}=13$ with $|t|$ facing $y$. Step 3. We have $f(h^{-1}(\frac{5}{12}))=sin(t)=\frac{5}{13}$