Chapter 7.1, Problem 79E

To determine

To verify the identity:

  $\frac{1}{\sec x+\tan x}+\frac{1}{\sec x-\tan x}=2\sec x$

Explanation of Solution

Given:

An identity is given as

  $\frac{1}{\sec x+\tan x}+\frac{1}{\sec x-\tan x}=2\sec x$

Concept Used:

The trigonometric identities:

  $\begin{array}{l}{\cos }^{2}x=1-{\sin }^{2}x\\ \sec x=\frac{1}{\cos x}\\ \tan x=\frac{\sin x}{\cos x}\end{array}$

Algebraic identity

  $a^2-b^2=\left(a-b\right)\left(a+b\right)$

Calculation:

In order to verify the identity:

  $\frac{1}{\sec x+\tan x}+\frac{1}{\sec x-\tan x}=2\sec x$

Take left hand of above identity and simplify it using the trigonometric identities.

As,

  $\begin{array}{c}\frac{1}{\sec x+\tan x}+\frac{1}{\sec x-\tan x}=\frac{\sec x-\tan x+\sec x+\tan x}{\left(\sec x+\tan x\right)\left(\sec x-\tan x\right)}\\ =\frac{2\sec x}{{\sec }^{2}x-{\tan }^{2}x}\\ =\frac{2\sec x}{1}\\ =2\sec x\\ =\text{Right}\text{hand}\end{array}$

Thus, identity is verified, i.e.

  $\frac{1}{\sec x+\tan x}+\frac{1}{\sec x-\tan x}=2\sec x$