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 659: 36


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Work Step by Step

$\frac{\csc x-\sec x}{\csc x+\sec x}$ Apply the reciprocal identity of trigonometry, $\csc x=\frac{1}{sinx}$ , and $\sec x=\frac{1}{\cos x}$. Then the above expression can be further simplified as: $\frac{\csc x-\sec x}{\csc x+\sec x}=\frac{\frac{1}{\sin x}-\frac{1}{\cos x}}{\frac{1}{\sin x}+\frac{1}{\cos x}}$ We multiply the numerator and denominator by $\cos x$. $\begin{align} & \frac{\frac{1}{\sin x}-\frac{1}{\cos x}}{\frac{1}{\sin x}+\frac{1}{\cos x}}=\frac{\frac{1}{\sin x}-\frac{1}{\cos x}}{\frac{1}{\sin x}+\frac{1}{\cos x}}.\frac{\cos x}{\cos x} \\ & =\frac{\frac{\cos x}{\sin x}-\frac{\cos x}{\cos x}}{\frac{\cos x}{\sin x}+\frac{\cos x}{\cos x}} \\ & =\frac{\frac{\cos x}{\sin x}-1}{\frac{\cos x}{\sin x}+1} \end{align}$ By using the quotient identity of trigonometry $\cot x=\frac{\cos x}{\sin x}$, the above expression can be further simplified as: $\frac{\frac{\cos x}{\sin x}-1}{\frac{\cos x}{\sin x}+1}=\frac{\cot x-1}{\cot x+1}$ Thus, the left side of the expression is equal to the right side, which is $\frac{\csc x-\sec x}{\csc x+\sec x}=\frac{\cot x-1}{\cot x+1}$.
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