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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.

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  • $\begingroup$ How can you re-write $\tan 2x$ in terms of $\tan x$? $\endgroup$ – Blue Apr 4 '19 at 15:34
  • $\begingroup$ What did you try? It's involves a straightforward simplification $\endgroup$ – pi-π Apr 4 '19 at 15:34
  • $\begingroup$ $\frac{\sin(2x)}{\cos(2x)}$. I know. I have tried. $\endgroup$ – Johnny Bueti Apr 4 '19 at 15:35
  • $\begingroup$ @Fehniix: $$\tan(a+b) = \frac{\tan a + \tan b}{1-\tan a \tan b}$$ so $\tan(x+x) = \cdots$? $\endgroup$ – Blue Apr 4 '19 at 15:38
  • $\begingroup$ @Blue had no clue there was such an equivalence! Thank you. $\endgroup$ – Johnny Bueti Apr 4 '19 at 15:43
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Defining $t := \tan x$ to save typing ... $$\begin{align} \tan 2x (1+\tan x) \cot x = \frac{2t}{1-t^2}\cdot(1+t)\cdot\frac{1}{t} = \frac{2t}{(1+t)(1-t)}\cdot(1+t)\cdot\frac{1}{t} = \frac{2}{1-t} \end{align}$$

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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 $\cos(x)$.

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  • $\begingroup$ It's actually easier just to write everything in terms of $\tan x$. $\endgroup$ – Blue Apr 4 '19 at 15:36
  • $\begingroup$ How so, Blue? Thank you guys. $\endgroup$ – Johnny Bueti Apr 4 '19 at 15:40
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$$\tan(2x) (1+\tan(x)) \cot(x) = \frac{2\sin(x)\cos(x)}{\cos^2(x)-\sin^2(x)}\left(\frac{\sin(x)+\cos(x)}{\cos(x)} \right)\frac{\cos(x)}{\sin(x)}$$

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

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

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