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: 91


The given statement does not make sense.

Work Step by Step

We have to simplify the identity $\frac{\left( \sec x+1 \right)\left( \sec x-1 \right)}{{{\sin }^{2}}x}$. The most efficient way is to multiply the numerator and then use the Pythagorean identity. And the numerator can be multiplied as shown below: $\frac{\left( \sec x+1 \right)\left( \sec x-1 \right)}{{{\sin }^{2}}x}=\frac{{{\sec }^{2}}x-1}{{{\sin }^{2}}x}$ Now, the Pythagorean identity, which is $1+{{\tan }^{2}}x={{\sec }^{2}}x$ can be further derived to ${{\tan }^{2}}x={{\sec }^{2}}x-1$ and can be used to solve the equation further. $\begin{align} & \frac{\left( \sec x+1 \right)\left( \sec x-1 \right)}{{{\sin }^{2}}x}=\frac{{{\tan }^{2}}x}{{{\sin }^{2}}x} \\ & \frac{\left( \sec x+1 \right)\left( \sec x-1 \right)}{{{\sin }^{2}}x}=\frac{\frac{{{\sin }^{2}}x}{{{\cos }^{2}}x}}{{{\sin }^{2}}x} \\ & \frac{\left( \sec x+1 \right)\left( \sec x-1 \right)}{{{\sin }^{2}}x}=\frac{1}{{{\cos }^{2}}x} \\ \end{align}$
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