Algebra and Trigonometry 10th Edition

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

Chapter 7 - Chapter Test - Page 554: 3



Work Step by Step

We know that $1+\tan^2 x=\sec^2 x $ $\dfrac{\sec^4 x -\tan^4 x}{\sec^2 x +\tan^2 x}=\dfrac{(\sec^2 x +\tan^2 x)(\sec^2 x -\tan^2 x)}{\sec^2 x +\tan^2 x}$ or, $=\sec^2 x -\tan^2 x$ or, $\dfrac{\sec^4 x -\tan^4 x}{\sec^2 x +\tan^2 x}=1$
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