Question

please answer of this question

Answer 1

What is that!!!!!???????????

Answer 2

Solution-

L.H.S

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

$=\frac{1\times (\sec x +\tan x)}{(\sec x -\tan x)\times (\sec x +\tan x)}-\frac{1}{\cos x}$

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

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

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

$=\frac{1+\sin x-1}{\cos x}$

$=\frac{\sin x}{\cos x}$

$=\tan x$

R.H.S

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

$=\frac{1}{\cos x}-\frac{1\times (\sec x -\tan x)}{(\sec x +\tan x)\times (\sec x -\tan x)}$

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

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

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

$=\frac{1-1+\sin x}{\cos x}$

$=\frac{\sin x}{\cos x}$

$=\tan x$

∴L.H.S = R.H.S