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$$\frac{\tan{(\frac{\pi}{4}+x)}-\tan{(\frac{\pi}{4}-x)}}{\tan{(\frac{\pi}{4}+x)}+\tan{(\frac{\pi}{4}-x)}} = 2\sin{x}\cos{x}$$

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On L.H.S, I've tried to write it using the sum and difference formula so it becomes

$$\frac{\dfrac{1+\tan x}{1-\tan x}-\dfrac{1-\tan x}{1+\tan x}}{\dfrac{1+\tan x}{1-\tan x}+\dfrac{1-\tan x}{1+\tan x}}$$

Then I try to rationalize and it got really messy $$\frac{\dfrac{(1+\tan x)(1+ \tan x)}{(1-\tan x)(1+\tan x)} - \dfrac{(1+\tan x)(1+\tan x)}{(1-\tan x)(1+\tan x)}}{\dfrac{(1+\tan x)(1+\tan x)}{(1-\tan x)(1+\tan x)} + \dfrac{(1+\tan x)(1+\tan x)}{(1-\tan x)(1+\tan x)}}$$

Somehow I distributed the terms and try to simplify and I got stuck. Any help would be appreciated. Thank you.

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You may write $$ \begin{align} \dfrac{\dfrac{1+\tan x}{1-\tan x}-\dfrac{1-\tan x}{1+\tan x}}{\dfrac{1+\tan x}{1-\tan x}+\dfrac{1-\tan x}{1+\tan x}}=\dfrac{(1+\tan x)^2-(1-\tan x)^2}{(1+\tan x)^2+(1-\tan x)^2} &=\dfrac{4\tan x}{2(1+\tan^2x)}\\\\&=2\tan x \cos^2x\\\\&=2\sin x\cos x \end{align} $$

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  • $\begingroup$ Thanks for replying, may I ask how did you arrive the third step from the second step? $\endgroup$ – Titration Apr 13 '15 at 0:24
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    $\begingroup$ @Titration Just expanding. $(a+b)^2-(a-b)^2=4ab$. Same for the denominator. Thanks. $\endgroup$ – Olivier Oloa Apr 13 '15 at 0:29
  • $\begingroup$ Can you also explain how did you go to the next step after expanding? $\endgroup$ – Titration Apr 13 '15 at 0:39
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    $\begingroup$ @Titration To understand what he did, observe that $$\frac{1}{1+\tan^2x}=\frac{1}{1+\frac{\sin^2x}{\cos^2x}}=\frac{\cos^2x}{\cos^2x+\sin^2x}=\frac{\cos^2x}{1}=\cos^2x$$ $\endgroup$ – P. Appell Apr 13 '15 at 6:41
  • $\begingroup$ @P.Appell thank you very much! I didn't know that I can rewrite $tan^2x$ like that $\endgroup$ – Titration Apr 13 '15 at 11:08
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multiplying the top and bottom by $\cos(\pi/4 - x)\cos(\pi/4+x)$ gives you $$\begin{align}\frac{\tan{(\frac{\pi}{4}+x)}-\tan{(\frac{\pi}{4}-x)}}{\tan{(\frac{\pi}{4}+x)}+\tan{(\frac{\pi}{4}-x)}} &= \frac{\sin(\pi/4+x)\cos(\pi/4-x)-\sin(\pi/4-x)\cos(\pi/4 + x)} {\sin (\pi/4+x)\cos(\pi/4-x) + \sin(\pi/4 - x)\cos(\pi/4 + x)}\\ &=\frac{\sin(\pi/4 + x-(\pi/4-x)}{\sin(\pi/4+x + \pi/4 - x) } \\ &= \sin 2x \\ &= 2\sin x \cos x\end{align}$$

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LHS = $\dfrac{(1+ \tan x )^2 - (1- \tan x )^2 }{(1+ \tan x )^2 + (1- \tan x )^2 }$

$=\dfrac{4 \tan x}{2(1 + tan^2 x)} $

can you take it from there ?

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