Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.4 Equations Involving Inverse Trigonometric Functions - 6.4 Exercises - Page 286: 36



Work Step by Step

Given: $cot^{-1}x=tan^{-1}\frac{4}{3}$ Consider $tan^{-1}\frac{4}{3}=P$ $tan P=\frac{4}{3}$ Apply trigonometric identity for tangent. $tanx=\frac{opp}{adj}=\frac{4}{3}$ Thus, $hyp=\sqrt {4^{2}+3^{2}}=\sqrt {16+9}=\sqrt {25}=5$ Likewise, $cotP=\frac{adj}{opp}=\frac{3}{4}$ $cot^{-1}x=P$ gives $x=cotP$ Hence, $x=\frac{3}{4}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.