Answer
$\frac{tan^2x}{1+sec~x}=sec~x-1$
Work Step by Step
$1+tan^2x=sec^2x$
$tan^2x=sec^2x-1=sec^2x-1^2=(sec~x+1)(sec~x-1)$
$\frac{tan^2x}{1+sec~x}=\frac{(sec~x+1)(sec~x-1)}{1+sec~x}=sec~x-1$
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: 16
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