Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

APPENDIX D - Trigonometry - D Exercises - Page A 32: 48


$tan^{2}\alpha-sin^{2}\alpha=tan^{2}\alpha$ $sin^{2}\alpha$

Work Step by Step

We need to prove the identity $tan^{2}\alpha-sin^{2}\alpha=tan^{2}\alpha$ $sin^{2}\alpha$ $tan^{2}\alpha-sin^{2}\alpha= \frac{sin^{2}\alpha}{cos^{2}\alpha}-sin^{2}\alpha$ $tan^{2}\alpha-sin^{2}\alpha=\frac{sin^{2}\alpha-sin^{2}\alpha cos^{2}\alpha}{cos^{2}\alpha}$ Thus, $tan^{2}\alpha-sin^{2}\alpha=\frac{1-cos^{2}}{cos^{2}\alpha}\times sin^{2}\alpha$ ${1-cos^{2}\alpha}=sin^{2}\alpha$ Now, $tan^{2}\alpha-sin^{2}\alpha=\frac{sin^{2}\alpha}{cos^{2}\alpha}\times sin^{2}\alpha$ Hence, $tan^{2}\alpha-sin^{2}\alpha=tan^{2}\alpha$ $sin^{2}\alpha$
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