Prove that $(G, *)$ is abelian group, for $ x * y = \tan^{-1}(tan(x) + tan(y))$ I have some troubles solving this problem. In order to prove that $(G, *)$ is an abelian group I have to find the identity element of the group, first;
$\exists \ e \in \ G \ and \ x \in G$ such that $x * e = e * x = x$
For this problem I have to prove that $x * e = tan^{-1}(tan(x) + tan(e)) = x$.
But I don't know where to begin, the expression of the binary function is pretty hard to simplify.
 A: Hint: you just need some $e$ with $\tan(e)=0$, then you have 
$x*e = \arctan ( \tan x + 0 ) = x$.
A: Let $e$ be the identity element, then
\begin{align*}
x * e & = x\\
\arctan(\tan(x) + \tan(e))& =x\\
 \tan(x) + \tan(e) & = \tan (x)\\
\tan(e)&=0.
\end{align*}
This suggests that $e=0$ can act as identity.
A: First, obviously, this operation is commutative on $\Bbb R\times\Bbb R $.
Second, also obviously, it is associative on $\Bbb R\times\Bbb R $.
You need to show that there exists a neutral element $e$; in other words,
$$\forall x\in  \Bbb R\quad \arctan (\tan x+ \tan e) = x,$$
which implies - after taking $\tan$ of both sides - that $\tan e =0$. Can you find such a $e$? After finding this $e$ showing that every element is invertible is a breeze.
A: It’s only true when $x$ and $y$ are restricted to the interval $\langle-\pi/2,\pi/2\rangle$. For when they are so restricted, $\tan$ and $\arctan$ become two-sided inverses of each other. Once you do that, you’ve just transformed ordinary addition on the open interval $\langle-\infty,\infty\rangle$ into the new addition on the bounded interval.
