Algebra and Trigonometry 10th Edition

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

Chapter 7 - Review Exercises - Page 552: 16



Work Step by Step

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