Trigonometry 7th Edition

Published by Cengage Learning
ISBN 10: 1111826854
ISBN 13: 978-1-11182-685-7

Chapter 5 - Section 5.1 - Proving Identities - 5.1 Problem Set - Page 279: 42

Answer

As left side transforms into right side, hence given identity- $\frac{\cos x}{1 + \sin x} - \frac{1 - \sin x}{\cos x}$ = $0$ is true.

Work Step by Step

Given identity is- $\frac{\cos x}{1 + \sin x} - \frac{1 - \sin x}{\cos x}$ = $0$ Taking L.S. $\frac{\cos x}{1 + \sin x} - \frac{1 - \sin x}{\cos x}$ = $\frac{\cos^{2} x - (1 - \sin x)(1 + \sin x) }{(1 + \sin x) \cos x} $ = $\frac{\cos^{2} x - (1^{2} - \sin^{2} x ) }{(1 + \sin x) \cos x} $ {Recall $(a-b)(a+b) $ = $a^{2} - b^{2}$ } = $\frac{\cos^{2} x - (1 - \sin^{2} x ) }{(1 + \sin x) \cos x} $ = $\frac{\cos^{2} x - (\cos^{2} x ) }{(1 + \sin x) \cos x} $ ( Recall first Pythagorean identity, $1 - \cos^{2} x =\sin^{2} x$ ) = $\frac{\cos^{2} x - \cos^{2} x }{(1 + \sin x) \cos x} $ = $\frac{0 }{(1 + \sin x) \cos x} $ = $0$ = R.S. As left side transforms into right side, hence given identity- $\frac{\cos x}{1 + \sin x} - \frac{1 - \sin x}{\cos x}$ = $0$ is true.
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