Question

Prove that : $\frac{1}{\sec x-\tan x}-\frac{1}{\sec x}=\tan x(1+\sin x)$

Answer 1

Now,

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

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

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

$=\frac{\tan x}{{\sec }^{2}x(1-\sin x)}$

$=\frac{\sin x}{\sec x(1-\sin x)}$

$=\frac{\sin x(1+\sin x)}{\sec x(1-{\sin }^{2}x)}$

$=\frac{\sin x(1+\sin x)}{\sec x.{\cos }^{2}x}$ [ Since $1-{\sin }^{2}x={\cos }^{2}x$]

$=\tan x(1+\sin x)$.