trigonometry - How to prove that $csc x - 2cos x cot 2x = 2 sin x$

My idea was...

$$\begin{align} & \text{LHS} \\[10pt] = {} & \frac 1 {\sin x} - \frac{2\cos x}{\tan2x} \\[10pt] = {} & \frac{\tan2x-\sin2x}{\sin x\tan2x} \end{align}$$

from here, I don't know how to continue, please help! thanks

ps, please teach me how to use the "divide" symbol

Answer

Somewhere you'll need a double-angle formula:

$$\tan(2x) = \frac{2\tan x}{1 - \tan^2 x}$$
or
$$\tan(2x) = \frac{\sin(2x)}{\cos(2x)} = \frac{2\sin x\cos x}{\cos^2 x - \sin^2 x}.$$