Answer
The identity is verified.
$sec^2x~cot~x-cot~x=tan~x$
Work Step by Step
$1+tan^2x=sec^2x~~$ (Pythagorean identity):
$tan^2x=sec^2x-1$
$cot~x=\frac{1}{tan~x}$
$sec^2x~cot~x-cot~x=(sec^2x-1)cot~x=tan^2x~\frac{1}{tan~x}=tan~x$
Algebra and Trigonometry 10th Edition
by Larson, Ron
Published by
Cengage Learning
ISBN 10:
9781337271172
ISBN 13:
978-1-33727-117-2
Chapter 7 - Review Exercises - Page 552: 20
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