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: 14



Work Step by Step

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