## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 6 - Section 6.1 - The Law of Sines - Exercise Set - Page 723: 80

#### Answer

See the explanation below.

#### Work Step by Step

Consider the given expression $\csc x{{\cos }^{2}}x+\sin x=\csc x$. Using the reciprocal property, we get $\csc x=\frac{1}{\sin x}$ \begin{align} & \csc x{{\cos }^{2}}x+\sin x=\frac{1}{\sin x}\cdot \frac{{{\cos }^{2}}x}{1}+\sin x \\ & =\frac{{{\cos }^{2}}x+{{\sin }^{2}}x}{\sin x}. \end{align} Now, we will use the property of ${{\sin }^{2}}x+{{\cos }^{2}}x=1$. This implies \begin{align} & \csc x{{\cos }^{2}}x+\sin x=\frac{1}{\sin x} \\ & =\csc x. \end{align} Hence, the provided expression is verified.

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