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: 66


$sec(arctan~3x)=±\sqrt {1+9x^2}$

Work Step by Step

Let $u=arctan~3x~~$ (Range: $-\frac{\pi}{2}\lt u\lt\frac{\pi}{2}$) Then: $3x=tan~u$ $sec^2u=1+tan^2u$ $sec^2u=1+(3x)^2=1+9x^2$ $sec~u=±\sqrt {1+9x^2}$ $sec(arctan~3x)=sec~u=±\sqrt {1+9x^2}$
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