Why is $\arctan(\tan(y))=y+2k \pi$? Why is $\arctan(\tan(y))=y+2k \pi$?

When the cycle of both $\arctan$ and $\tan$ are $\pi$?
It's actually wrong. The correct answer is actually
$$\arctan(\tan(y))=y+k\pi\quad k\in\mathbb{Z}\text{ s.t. } |y+k\pi|&lt;\pi/2$$

if the standard $\arctan:\mathbb{R}\to(-\pi/2,\pi/2)$ function is used as an inverse.
because the function tan(x) have period $2\pi$.
I edited the answer.