Answer
$\frac{u}{\sqrt {1-u^2}}$
Work Step by Step
1. Let $sin^{-1}u=t, -\frac{\pi}{2}\le t\le \frac{\pi}{2}$, we have $sin(t)=u$.
2. Let $y=u,r=1$, we have $x=\sqrt {1-u^2}$
3. Thus $tan(sin^{-1}u)=tan(t)=\frac{y}{x}=\frac{u}{\sqrt {1-u^2}}$
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: 59
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