Answer
$cos~x(sec~x-cos~x)=sin^2x$
Work Step by Step
We know that:
$sec~x=\frac{1}{cos~x}$
$sin^2x+cos^2x=1$
$sin^2x=1-cos^2x$
So:
$cos~x(sec~x-cos~x)=cos~x(\frac{1}{cos~x}-cos~x)=1-cos^2x=sin^2x$
Algebra and Trigonometry 10th Edition
by Larson, Ron
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: 36
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