Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 4 - Section 4.7 - Inverse Trigonometric Functions - Exercise Set - Page 627: 63


$\dfrac{\sqrt {1-x^2}}{x}$

Work Step by Step

Suppose $\theta =\cos^{-1} x $ This gives: $ x =\cos \theta $ Since, $ r=\sqrt {a^2+y^2} \implies a=\sqrt {r^2-y^2}$ This implies that $ a=\sqrt {1^2-x^2}=\sqrt {1-x^2}$ Therefore, we have $\tan(\cos^{-1} x)=\dfrac{\sqrt {1-x^2}}{x}$
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