How do I prove that 2 cot 2x = cot x-tan x?

We will start off working on the right hand side.

cot x - tan x

= [cos x / sin x] - [sin x / cos x]

= [ (cos x) ^ 2 - (sin x) ^2] / [sin x cos x]

This is where it gets a bit tougher if you do not have your formula list with you.

(cos x) ^ 2 - (sin x) ^2 = cos (2x)

sin 2x = 2 sin x cos x

Note that by arranging the second formula, we will have sin x cos x = (1/2) sin 2x

Hence, we will get:

[ (cos x) ^ 2 - (sin x) ^2] / [sin x cos x]

= [cos 2x] / (1/2) [sin 2x]

= 2[cos 2x] / [sin 2x]

= 2cot 2x