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: 65


The solution of the given question is $2\sec x$.

Work Step by Step

Let us consider the left side of the given expression: $\frac{1}{\sec x+\tan x}+\frac{1}{\sec x-\tan x}$ By using the formula ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\times \left( a-b \right)$ , the above expression can be further simplified as: $\frac{1}{\sec x+\tan x}+\frac{1}{\sec x-\tan x}=\frac{{{\sec }^{2}}x-{{\tan }^{2}}x}{\sec x+\tan x}+\frac{{{\sec }^{2}}x-{{\tan }^{2}}x}{\sec x-\tan x}$ Now, the above equation is further simplified in order to eliminate the denominator $\begin{align} & \frac{{{\sec }^{2}}x-{{\tan }^{2}}x}{\sec x+\tan x}+\frac{{{\sec }^{2}}x-{{\tan }^{2}}x}{\sec x-\tan x}=\frac{\left( \sec x+\tan x \right)\left( \sec x-\tan x \right)}{\sec x+\tan x} \\ & +\frac{\left( \sec x+\tan x \right)\left( \sec x-\tan x \right)}{\sec x-\tan x} \\ & =\sec x-\tan x+\sec x+\tan x \\ & =2\sec x \end{align}$ Conjecture: Left side is equal to $2\sec x$. Thus, the left side of the expression is equal to the right side, which is $\frac{1}{\sec x+\tan x}+\frac{1}{\sec x-\tan x}=2\sec x$.
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