Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 5 - Section 5.1 - Verifying Trigonometric Identities - Exercise Set - Page 660: 67


The expression in terms of the function $\cos x$ is $\frac{1}{\cos x}$.

Work Step by Step

In order to write the above expression in terms of the function $\cos x$ , the following course of action needs to be followed as, $\frac{\tan x+\cot x}{\csc x}=\frac{\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}}{\frac{1}{\sin x}}$ Now, the basic rule of reversing the fraction is to be followed by reversing the denominator and the numerator: $\begin{align} & \left( \frac{\sin x}{\cos x}+\frac{\cos x}{\sin x} \right)\frac{\sin x}{1}=\frac{{{\sin }^{2}}x}{\cos x}+\frac{\sin x\cos x}{\sin x} \\ & =\frac{1-{{\cos }^{2}}x}{\cos x}+\frac{\cos x}{1} \\ & =\frac{1-{{\cos }^{2}}x}{\cos x}+\frac{{{\cos }^{2}}x}{\cos x} \\ & =\frac{1}{\cos x} \end{align}$ Therefore, the $\frac{\tan x+\cot x}{\csc x}$ expression is written in the terms of the function $\cos x$ as $\frac{1}{\cos x}$. Hence, the expression in terms of the functions $\cos x$ is $\frac{1}{\cos x}$.
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