How we can prove this identity $\arctan(\tan x)=x+n\pi$

Question

Generally $\arctan(\tan x)=x+n\pi$ for some $n\in \mathbb{Z}$, and $n=0$ if and only if $x \in (-\frac{\pi}{2},\frac{\pi}{2})$.

How we can prove this identity.

Answer 1

It is a simple consequence of the definition of the $\arctan$ function. We have:

$y = \arctan(z) \Rightarrow z = \tan(y)$

so

$y = \arctan(\tan(x)) \Rightarrow \tan(x) = \tan(y)$

so one solution is $y=x$ and the other solutions come from the fact that the $\tan$ function is periodic with period $\pi$.

Answer 2

Note that $\tan(x)$ has a period of $\pi$. That is to say, $$\tan(x+n\pi)\equiv\tan(x)\space \forall n\in\Bbb Z$$

Taking the $\arctan$ of both sides tells us $$x+n\pi=\arctan(\tan(x))$$ is indeed acceptable.