Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Appendix D - Trigonometry - D Exercises: 52

Answer

$\frac{1}{1-sin\theta}+\frac{1}{1+sin\theta}=2sec^{2}\theta$

Work Step by Step

Need to prove the identity $\frac{1}{1-sin\theta}+\frac{1}{1+sin\theta}=2sec^{2}\theta$ Let us take left side solve first. $\frac{1}{1-sin\theta}+\frac{1}{1+sin\theta}=\frac{1+sin\theta}{(1-sin\theta)({1+sin\theta)}}+\frac{1-sin\theta}{(1-sin\theta)({1+sin\theta)}}$ $\frac{1}{1-sin\theta}+\frac{1}{1+sin\theta}=\frac{2}{1-sin^{2}\theta}$ Since, $sin^{2}\theta+cos^{2}\theta=1$ $\frac{1}{1-sin\theta}+\frac{1}{1+sin\theta}=\frac{2}{cos^{2}\theta}$ But $\frac{1}{cos\theta}=sec\theta$ Hence, $\frac{1}{1-sin\theta}+\frac{1}{1+sin\theta}=2sec^{2}\theta$
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