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 628: 123


$\tan^{-1} (\dfrac{33}{x})- \tan^{-1} (\dfrac{8}{x})$

Work Step by Step

Using problem 93, we get: Suppose $\tan (\alpha+\theta)=\dfrac{33}{x}$ This gives: $\alpha+\theta= \tan^{-1} \dfrac{33}{x}$ And also: $\tan \alpha=\dfrac{8}{x}$ Which gives: $\alpha= \tan^{-1} \dfrac{8}{x}$ Therefore, we have $\theta=\alpha+\theta-\alpha= \tan^{-1} (\dfrac{33}{x})- \tan^{-1} (\dfrac{8}{x})$
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