plane trigonometry

Prove the identity
$\dfrac{2\sin x - 2\cos x}{\tan x - \cot x} = \dfrac{\sin 2x}{\sin x + \cos x}$

$\dfrac{2(\sin x - \cos x)}{\dfrac{\sin x}{\cos x} - \dfrac{\cos x}{\sin x}} = \dfrac{\sin 2x}{\sin x + \cos x}$

$\dfrac{2(\sin x - \cos x)}{\dfrac{\sin^2 x- \cos^2 x}{\sin x \, \cos x}} = \dfrac{\sin 2x}{\sin x + \cos x}$

$\dfrac{2(\sin x - \cos x)\sin x \, \cos x}{\sin^2 x- \cos^2 x} = \dfrac{\sin 2x}{\sin x + \cos x}$

$\dfrac{2(\sin x - \cos x)\sin x \, \cos x}{(\sin x - \cos x)(\sin x + \cos x)} = \dfrac{\sin 2x}{\sin x + \cos x}$

$\dfrac{2\sin x \, \cos x}{\sin x + \cos x} = \dfrac{\sin 2x}{\sin x + \cos x}$

$\dfrac{\sin 2x}{\sin x + \cos x} = \dfrac{\sin 2x}{\sin x + \cos x}$