## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 11 - Section 11.4 - Introduction to Derrivatives - Exercise Set - Page 1177: 84

#### Answer

See the verification below.

#### Work Step by Step

Consider the identity: $\csc x-\cos x\cot x=\sin x$ Solve the left side of the above identity, \begin{align} & \csc x-\cos x\cot x=\frac{1}{\sin x}-\cos x\frac{\cos x}{\sin x} \\ & =\frac{1}{\sin x}-\frac{{{\cos }^{2}}x}{\sin x} \\ & =\frac{1-{{\cos }^{2}}x}{\sin x} \end{align} Now, use the identity $1-{{\cos }^{2}}x={{\sin }^{2}}x$ and solve further, \begin{align} & \csc x-\cos x\cot x=\frac{{{\sin }^{2}}x}{\sin x} \\ & =\sin x \end{align} Now, the right side of the identity $\csc x-\cos x\cot x=\sin x$ is also $\sin x$ Thus, the left side and the right side of the identity $\csc x-\cos x\cot x=\sin x$ are equal. Hence, the identity $\csc x-\cos x\cot x=\sin x$ is verified.

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