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: 70


$\dfrac{\sqrt 2}{x}$

Work Step by Step

Suppose $\theta =\tan^{-1} (\dfrac{x}{\sqrt 2})$ This gives: $\tan \theta=\dfrac{x}{\sqrt 2}$ Since, $ r=\sqrt {x^2+y^2}$ We know $\cot \theta=\dfrac{1}{\tan \theta}=\dfrac{\sqrt 2}{x}$ Therefore, we have $\cot [\tan^{-1} (\dfrac{x}{\sqrt 2})]=\dfrac{\sqrt 2}{x}$
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