Precalculus (10th Edition)

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

Chapter 8 - Applications of Trigonometric Functions - 8.1 Right Triangle Trigonometry; Applications - 8.1 Assess Your Understanding - Page 519: 26



Work Step by Step

We know that $\sin{(\theta)}=\cos{(90^{o}-\theta)}$, and $\cot{(\theta)}=\frac{\sin{(\theta)}}{\cos{(\theta)}}$, hence $\cot{(40^{o})}-\frac{\sin{(50^{o})}}{\sin{(40^{o})}}=\cot{(40^{o})}-\frac{\cos{(40^{o})}}{\sin{(40^{o})}}=\cot{(40^{o})}-\cot{(40^{o})}=0.$
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.