Trigonometry (11th Edition) Clone

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

Chapter 5 - Review Exercises - Page 249: 62


$sec^2~\alpha-1 = \frac{sec~2\alpha-1}{sec~2\alpha+1}$

Work Step by Step

We can verify that the equation is an identity: $\frac{sec~2\alpha-1}{sec~2\alpha+1}$ $=\frac{\frac{1}{cos~2\alpha}-1}{\frac{1}{cos~2\alpha}+1}$ $=\frac{\frac{1-cos~2\alpha}{cos~2\alpha}}{\frac{1+cos~2\alpha}{cos~2\alpha}}$ $=\frac{1-cos~2\alpha}{1+cos~2\alpha}$ $=\frac{1-(2cos^2~\alpha-1)}{1+(2cos^2~\alpha-1)}$ $= \frac{1-2cos^2~\alpha+1}{1+2cos^2~\alpha-1}$ $= \frac{2-2cos^2~\alpha}{2cos^2~\alpha}$ $= \frac{2}{2cos^2~\alpha}- \frac{2cos^2~\alpha}{2cos^2~\alpha}$ $= \frac{1}{cos^2~\alpha}-1$ $= sec^2~\alpha-1$
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