Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.1 Inverse Circular Functions - 6.1 Exercises - Page 266: 96


$tan~(arccos~u) = \frac{\sqrt{1-u^2}}{u}$

Work Step by Step

Let $~~\theta = arccos~u$ Then $~~cos~\theta = u$ $tan~\theta = \frac{sin~\theta}{cos~\theta}$ $tan~\theta = \frac{\sqrt{1-cos^2~\theta}}{cos~\theta}$ $tan~\theta = \frac{\sqrt{1-u^2}}{u}$ Therefore, $~~tan~(arccos~u) = \frac{\sqrt{1-u^2}}{u}$
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