Trigonometry (11th Edition) Clone

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

Chapter 5 - Trigonometric Identities - Section 5.2 Verifying Trigonometric Identities - 5.2 Exercises - Page 208: 21


$$\frac{1}{1+\cos x}-\frac{1}{1-\cos x}=-2\cos x\csc^2 x$$

Work Step by Step

$$A=\frac{1}{1+\cos x}-\frac{1}{1-\cos x}$$ $$A=\frac{1-\cos x}{(1+\cos x)(1-\cos x)}-\frac{1+\cos x}{(1+\cos x)(1-\cos x)}$$ $$A=\frac{1-\cos x-(1+\cos x)}{(1+\cos x)(1-\cos x)}$$ $$A=\frac{1-\cos x-1-\cos x}{(1+\cos x)(1-\cos x)}$$ $$A=\frac{-2\cos x}{(1+\cos x)(1-\cos x)}$$ $$A=\frac{-2\cos x}{1-\cos^2 x}$$ (for $a^2-b^2=(a-b)(a+b)$) - Pythagorean Identity: $$\sin^2 x=1-\cos^2 x$$ Therefore, $$A=\frac{-2\cos x}{\sin^2 x}$$ $$A=-2\cos x(\frac{1}{\sin x})^2$$ - Reciprocal Identity: $$\csc x=\frac{1}{\sin x}$$ So, $$A=-2\cos x\csc^2 x$$
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