Chapter 7 - Analytic Trigonometry - 7.2 The Inverse Trigonometric Functions (Continued) - 7.2 Assess Your Understanding - Page 458: 61

$\frac{\sqrt {1-u^2}}{|u|}$

Step 1. Letting $sec^{-1}u=t$, we have $sec(t)=u$ and $0\le t \le \pi$ Step 2. Form a right triangle with sides $x=1,r=|u|,y=\sqrt {u^2-1}$ and angle $|t|$ facing $y$. Step 3. As $sin(t)$ is positive in the above interval, we have $sin(sec^{-1}u)=sin(t)=\frac{\sqrt {1-u^2}}{|u|}$