Answer
$\frac{\sqrt {1-u^2}}{u}$
Work Step by Step
1. Let $cos^{-1}u=t, 0\le t\le\pi$, we have $cos(t)=u$.
2. Let $x=u,r=1$, we have $y=\sqrt {1-u^2}$
3. Thus $tan(cos^{-1}u)=tan(t)=\frac{y}{x}=\frac{\sqrt {1-u^2}}{u}$
Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)
by Sullivan III, Michael
Published by
Pearson
ISBN 10:
0-32193-104-1
ISBN 13:
978-0-32193-104-7
Chapter 6 - Analytic Trigonometry - Section 6.2 The Inverse Trigonometric Functions (Continued) - 6.2 Assess Your Understanding - Page 481: 60
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