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 626: 13



Work Step by Step

Suppose $\theta = \tan^{-1} (\dfrac{\sqrt 3}{3})$ This gives: $\tan \theta= \dfrac{\sqrt 3}{3}$ Now, $\theta = \tan^{-1} (\dfrac{\sqrt 3}{3})=\dfrac{\pi}{6}$ Therefore, $\tan^{-1} (\dfrac{\sqrt 3}{3})=\dfrac{\pi}{6}$
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.