Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 6 - 6.6 - Inverse Trigonometric Functions - 6.6 Exercises - Page 485: 69


$tan(arccos\frac{x}{3})=\frac{\sqrt {9-x^2}}{x}$

Work Step by Step

Let $u=arccos\frac{x}{3}~~$ (Range: $0\lt u\lt\pi$) Then: $\frac{x}{3}=cos~u$ $sin^2u+cos^2u=1$ $sin^2u=1-cos^2u$ $sin^2u=1-(\frac{x}{3})^2=1-\frac{x^2}{9}=\frac{9-x^2}{9}$ $sin~u=\frac{\sqrt {9-x^2}}{3}~~$ ($0\lt u\lt\pi$) $tan(arccos\frac{x}{3})=tan~u=\frac{sin~u}{cos~u}=\frac{\frac{\sqrt {9-x^2}}{3}}{\frac{x}{3}}=\frac{\sqrt {9-x^2}}{x}$
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