Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 5 - Test - Page 250: 13


$tan^2~x-sin^2~x = (tan~x~sin~x)^2$

Work Step by Step

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