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 487: 130c


The identity is verified. $arctan~x+arctan\frac{1}{x}=\frac{\pi}{2}$

Work Step by Step

Let $y=arctan~x$ $x=tan~y$ $\frac{1}{x}=cot~y=tan(\frac{\pi}{2}-y)$ $arctan~x+arctan\frac{1}{x}=arctan~x+arctan[tan(\frac{\pi}{2}-y)]=y+\frac{\pi}{2}-y=\frac{\pi}{2}$
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