Question

What is the expression $\frac{\tan x}{1+\sec x}-\frac{\tan x}{1-\sec x}$ equal to?

• A. $\text{cosec}x$
• B. $2\text{cosec}x$
• C. $2\sin x$
• D. $2\cos x$

Answer 1

The correct option is B $2\text{cosec}x$

$\frac{\tan x}{1+\sec x}-\frac{\tan x}{1-\sec x}$
$=\frac{\sin x}{1+\cos x}-\frac{\sin x}{\cos x-1}$
$=\sin x\left[\frac{\cos x-1-1\cos x}{{\cos }^{2}x-1}\right]$
$=\sin x\times \frac{2}{{\sin }^{2}x}=2\text{cosec}x$