2014/103) prove that $\cot 2x - \tan 2x = 2\cot 4x$

LHS = $\dfrac{\cos 2x }{\ sin 2x}- \dfrac{\sin 2x}{\cos 2x}$
= $\dfrac{\cos^2 2x - \sin^2 2x}{\sin 2x\cos 2x}$
= $\dfrac{\cos 4x}{\sin 2x\cos 2x}$
= $\dfrac{2\cos 4x}{2\sin 2x\cos 2x}$
= $\dfrac{2 \cos 4x}{\sin 4x}$
= $2 \cot 4x$