Question

State true or false.

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

• A. True
• B. False

Answer 1

The correct option is A True

Given

LHS

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

$⟹\frac{\sec x+\tan x+\sec x-\tan x}{{\sec }^{2}x-{\tan }^{2}x}$

$⟹2\sec x$

$⟹\frac{2}{\cos x}$