Find the value of $$\frac{\cot x - \tan x}{\cot 2x}$$​ is?

Correct option is A

Given:

$$\frac{\cot x - \tan x}{\cot 2x}$$

Formula used:

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

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

​​​​$$\sin2x = 2 \sin x \cos x$$

$${\cos^2 x - \sin^2 x} = \cos2x$$​​

Solution:

$$\frac{\cot x - \tan x}{\cot 2x}$$

=  $$\frac{\left( \frac{\cos x}{\sin x} - \frac{\sin x}{\cos x}\right)}{\left( \frac{\cos 2x}{\sin x2}\right)}$$

= $$\frac{\left( \frac{\cos^2 x - \sin^2 x}{\sin x \cos x}\right)}{\left( \frac{\cos 2x}{\sin x2}\right)}$$​​  

= ​​$${\left( \frac{\cos2x}{\sin x \cos x}\right)}\times{\left( \frac{\sin 2x}{\cos 2x}\right)}$$

= $$\frac{\sin 2x}{\sin x \cos x}$$

= $$\frac{2\sin x \cos x}{\sin x \cos x}$$

= 2

Thus, the correct answer is (a).