Chapter 6 - Analytic Trigonometry - Section 6.2 The Inverse Trigonometric Functions (Continued) - 6.2 Assess Your Understanding - Page 481: 63

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

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