Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 5 - Test - Page 244: 14

Answer

$\frac{tan~x-cot~x}{tan~x+cot~x} = 2sin^2~x-1$

Work Step by Step

$\frac{tan~x-cot~x}{tan~x+cot~x} = \frac{\frac{sin~x}{cos~x}-\frac{cos~x}{sin~x}}{\frac{sin~x}{cos~x}+\frac{cos~x}{sin~x}}$ $\frac{tan~x-cot~x}{tan~x+cot~x} = \frac{\frac{sin^2~x-cos^2~x}{sin~x~cos~x}}{\frac{sin^2~x+cos^2~x}{sin~x~cos~x}}$ $\frac{tan~x-cot~x}{tan~x+cot~x} = \frac{sin^2~x-cos^2~x}{sin^2~x+cos^2~x}$ $\frac{tan~x-cot~x}{tan~x+cot~x} = \frac{sin^2~x-cos^2~x}{1}$ $\frac{tan~x-cot~x}{tan~x+cot~x} = sin^2~x-(1-sin^2~x)$ $\frac{tan~x-cot~x}{tan~x+cot~x} = 2sin^2~x-1$
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.