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!

  • 2
    $\begingroup$ Do you know the addition formula for $\tan$? $\endgroup$ Mar 28, 2014 at 22:11
  • $\begingroup$ @DanielFischer, I don't think so... $\endgroup$
    – CS1
    Mar 28, 2014 at 22:12
  • 4
    $\begingroup$ $$\tan(a\pm b) = \frac{\tan(a)\pm\tan(b)}{1\mp\tan(a)\tan(b)}.$$ $\endgroup$ Mar 28, 2014 at 22:13
  • $\begingroup$ Thank you!!! Both of you help me a lot!! $\endgroup$
    – CS1
    Mar 28, 2014 at 22:14
  • 1
    $\begingroup$ You're very welcome :) $\endgroup$ Mar 28, 2014 at 22:15

1 Answer 1


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, $$\tan(\arctan x-\pi/4)=\dfrac{\tan \arctan x-\tan \pi/4}{1+\tan \arctan x\tan \pi/4}=\dfrac{x-1}{1+x}$$


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