The value of

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

$=\frac{\frac{1}{\tan x}-\tan x}{\frac{1}{\tan 2 x}}$

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

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

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

Therefore,

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