Chapter 6 - Analytic Trigonometry - Section 6.2 The Inverse Trigonometric Functions (Continued) - 6.2 Assess Your Understanding - Page 481: 61

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

1. Let $sec^{-1}u=t, 0\le t\le\pi, t\ne\frac{\pi}{2}$, we have $sec(t)=u, cos(t)=\frac{1}{u}$. 2. Let $x=1,r=|u|$, we have $y=\sqrt {u^2-1}$ 3. Thus $sin(sec^{-1}u)=sin(t)=\frac{y}{r}=\frac{\sqrt {u^2-1}}{|u|}$