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

Question

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

When the cycle of both $\arctan$ and $\tan$ are $\pi$?

Answer 1

because the function tan(x) have period $2\pi$. I edited the answer.

Answer 2

It's actually wrong. The correct answer is actually $$\arctan(\tan(y))=y+k\pi\quad k\in\mathbb{Z}\text{ s.t. } |y+k\pi|<\pi/2$$

if the standard $\arctan:\mathbb{R}\to(-\pi/2,\pi/2)$ function is used as an inverse.