Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 7 - Analytic Trigonometry - 7.2 The Inverse Trigonometric Functions (Continued) - 7.2 Assess Your Understanding - Page 458: 66

Answer

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

Work Step by Step

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 $x=|u|,r=1,y=\sqrt {1-u^2}$. Step 3. As $tan(t)$ and $sec(t)$ have the same sign in the above interval, we have $tan(sec^{-1}u)=tan(t)=\frac{\sqrt {1-u^2}}{u}$
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