Proving that $ \tan 2x \cdot (1 + \tan x) \cdot \cot x = \frac{2}{1 - \tan(x)} $

Question

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.

Answer 1

Defining $t := \tan x$ to save typing ... $$\begin{align} \tan 2x (1+\tan x) \cot x = \frac{2t}{1-t^2}\cdot(1+t)\cdot\frac{1}{t} = \frac{2t}{(1+t)(1-t)}\cdot(1+t)\cdot\frac{1}{t} = \frac{2}{1-t} \end{align}$$

Answer 2

It is true.

Write everything in terms of $\sin(x)$ and $\cos(x)$, and cancel common factors from numerator and denominator. You should end up with $$ \frac{2 \cos(x)}{\cos(x)-\sin(x)}$$ and then divide numerator and denominator by $\cos(x)$.

Answer 3

$$\tan(2x) (1+\tan(x)) \cot(x) = \frac{2\sin(x)\cos(x)}{\cos^2(x)-\sin^2(x)}\left(\frac{\sin(x)+\cos(x)}{\cos(x)} \right)\frac{\cos(x)}{\sin(x)}$$

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. $$\tan(2x) (1+\tan(x)) \cot(x) =\frac{2}{1-\tan(x)}$$