# 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.

• How can you re-write $\tan 2x$ in terms of $\tan x$? – Blue Apr 4 '19 at 15:34
• What did you try? It's involves a straightforward simplification – pi-π Apr 4 '19 at 15:34
• $\frac{\sin(2x)}{\cos(2x)}$. I know. I have tried. – Johnny Bueti Apr 4 '19 at 15:35
• @Fehniix: $$\tan(a+b) = \frac{\tan a + \tan b}{1-\tan a \tan b}$$ so $\tan(x+x) = \cdots$? – Blue Apr 4 '19 at 15:38
• @Blue had no clue there was such an equivalence! Thank you. – Johnny Bueti Apr 4 '19 at 15:43

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}

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 $$\cos(x)$$.

• It's actually easier just to write everything in terms of $\tan x$. – Blue Apr 4 '19 at 15:36
• How so, Blue? Thank you guys. – Johnny Bueti Apr 4 '19 at 15:40

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

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

• Wow. Thank you. – Johnny Bueti Apr 4 '19 at 15:39