Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 6 - Trigonometric Functions - 6.3 Properties of the Trigonometric Functions - 6.3 Assess Your Understanding - Page 395: 82



Work Step by Step

RECALL: $\cot \alpha =\dfrac {\cos \alpha }{\sin \alpha }$ Thus, $\cot \alpha -\dfrac {\cos \alpha }{\sin \alpha }=\cot{\alpha}-\cot{\alpha}=0$ Therefore, $\cot 20^o-\dfrac {\cos 20^o}{\sin 20^o}=0$
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.