Algebra and Trigonometry 10th Edition

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: 17


$\frac{sec^2x-1}{sin^2x}=sec^2x$ (e)

Work Step by Step

Use the Pythagorean Identity: $tan^2x+1=sec^2x$ $tan^2x=sec^2x-1$ And: $tan~x=\frac{sin~x}{cos~x}$ $\frac{sec^2x-1}{sin^2x}=\frac{tan^2x}{\frac{sin^2x}{1}}=\frac{sin^2x}{cos^2x}\frac{1}{sin^2x}=\frac{1}{cos^2x}=sec^2x$
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