Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 7 - 7.1 - Using Fundamental Identities - 7.1 Exercises - Page 513: 18


$\frac{cos^2(\frac{\pi}{2}-x)}{cos~x}=tan~x~sin~x$ (d)

Work Step by Step

$cos(\frac{\pi}{2}-x)=sin~x$ $tan~x=\frac{sin~x}{cos~x}$ $sec~x=\frac{1}{cos~x}$ $\frac{cos^2(\frac{\pi}{2}-x)}{cos~x}=\frac{sin^2x}{cos~x}=\frac{sin~x}{cos~x}sin~x=tan~x~sin~x$
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