Proving that $ \tan 2x \cdot (1 + \tan x) \cdot \cot x = \frac{2}{1 - \tan(x)} $ 
Given the following expression,
  $$ \tan(2x) \cdot (1 + \tan(x)) \cdot \cot(x) $$
  the exercise asks to simplify the expression and
  $$ \frac{2}{1 - \tan(x)} $$
  should be the simplified expression.

I have tried everything I possibly could, including letting WolframAlpha eat it to show alternative forms of the expression – nothing worked. 
What do you think? How could I go about simplifying this expression? Thank you.

$$\tan(2x) (1+\tan(x)) \cot(x) = \frac{2\sin(x)\cos(x)}{\cos^2(x)-\sin^2(x)}\left(\frac{\sin(x)+\cos(x)}{\cos(x)} \right)\frac{\cos(x)}{\sin(x)}$$
Simplifying you get
$$\tan(2x) (1+\tan(x)) \cot(x) = \frac{2(\sin(x)+\cos(x))\cos(x)}{(\cos(x)-\sin(x))(\cos(x)+\sin(x))} = \frac{2\cos(x)}{\cos(x)-\sin(x)}$$
i.e.
$$\tan(2x) (1+\tan(x)) \cot(x) =\frac{2}{1-\tan(x)}$$
