# Someone can explain me why $\tan(-\frac{\pi}4+\arctan x)=\frac{x-1}{x+1}$

Someone can explain me why $tan(-\frac{\pi}4+\arctan x)=\frac{x-1}{x+1}$??

I try to understand it, bot I don't understand how to came from one side to the other...

Thank you!

• Do you know the addition formula for $\tan$? Mar 28, 2014 at 22:11
• @DanielFischer, I don't think so...
– CS1
Mar 28, 2014 at 22:12
• $$\tan(a\pm b) = \frac{\tan(a)\pm\tan(b)}{1\mp\tan(a)\tan(b)}.$$ Mar 28, 2014 at 22:13
• Thank you!!! Both of you help me a lot!!
– CS1
Mar 28, 2014 at 22:14
• You're very welcome :) Mar 28, 2014 at 22:15

We know that $$\tan(a-b)=\dfrac{\tan a-\tan b}{1+\tan a\tan b}$$ So, let $b=\pi/4,a=\arctan x$. Then, $$\tan(\arctan x-\pi/4)=\dfrac{\tan \arctan x-\tan \pi/4}{1+\tan \arctan x\tan \pi/4}=\dfrac{x-1}{1+x}$$