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: 46

Answer

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

Work Step by Step

Given identity is- $\frac{\csc x - 1}{\csc x + 1} $ = $ \frac{1 - \sin x}{1 + \sin x}$ Taking L.S. $\frac{\csc x - 1}{\csc x + 1} $ = $\frac{\frac{1}{\sin x} - 1 }{\frac{1}{\sin x} + 1 } $ (using reciprocal identity) = $\frac{\frac{1}{\sin x} - \frac{\sin x}{\sin x} } {\frac{1}{\sin x} + \frac{\sin x}{\sin x}} $ = $\frac{\frac{1 -\sin x}{\sin x} }{\frac{1 +\sin x}{\sin x} } $ = $\frac{(1 -\sin x)\sin x} {(1 +\sin x)\sin x} $ = $ \frac{1 - \sin x}{1 + \sin x}$ = R.S. As left side transforms into right side, hence given identity- $\frac{\csc x - 1}{\csc x + 1} $ = $ \frac{1 - \sin x}{1 + \sin x}$ is true.
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