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: 37

Answer

As left side transforms into right side, hence given identity- $\csc B - \sin B$ = $\cot B \cos B$ is true.

Work Step by Step

Given identity is- $\csc B - \sin B$ = $\cot B \cos B$ Taking L.S. $\csc B - \sin B$ = $\frac{1}{\sin B} - \sin B$ ( Using ratio identity) = $\frac{1}{\sin B} - \sin B . \frac{\sin B}{\sin B}$ = $\frac{1}{\sin B} - \frac{\sin^{2} B}{\sin B}$ = $\frac{1 - \sin^{2} B}{\sin B}$ = $\frac{\cos^{2} B}{\sin B}$ ( From first Pythagorean identity, $1 - \sin^{2}\theta$ = $\cos^{2}\theta$) = $\frac{\cos B}{\sin B} . \cos B$ =$\cot B \cos B$ = R.S. As left side transforms into right side, hence given identity- $\csc B - \sin B$ = $\cot B \cos B$ is true.
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