# Thread: help me prove it

1. ## help me prove it

hello
let $\left ( a,x \right )\in \mathbb{R}^{2}$ such that $ax\neq 1$,prove the following ;
$arctan(x)+arctan(a)=arctan(\frac{a+x}{1-ax})+\epsilon \pi$
( $\epsilon \in \left \{ 1,-1,0 \right \}$ )
just hints guys,thanks a lot.

2. Hint #1:

$\frac{\tan(\arctan a)+\tan(\arctan x)}{1-\tan(\arctan a)\tan(\arctan x)}\ =\ \frac{a+x}{1-ax}\ =\ \frac{\tan\left[\arctan\left(\dfrac{a+x}{1-ax}\right)\right]+\tan(\epsilon\pi)}{1-\tan\left[\arctan\left(\dfrac{a+x}{1-ax}\right)\right]\tan(\epsilon\pi)}$ where $\epsilon\in\mathbb Z$

Hint #2:

$-\frac\pi2\,<\,\arctan\left(\frac{a+x}{1-ax}\right)\,<\,\frac\pi2$ and $-\pi\,<\,\arctan a+\arctan x\,<\,\pi$ (this is to get the range of values of $\epsilon)$

3. Hello, Raoh!

A slight variation of The Abstractionist's proof . . .

Let $\left(x,y \right)\in \mathbb{R}^{2}$ such that $xy \neq 1$, prove the following:

$\arctan(x)+\arctan(y)\:=\:\arctan\left(\frac{x+y}{ 1-xy}\right)+n\pi, \quad
n \in\{-1,0,1\}$
Let: . $\begin{array}{ccccc}\alpha \:=\:\arctan(x) &\Rightarrow& x \:=\:\tan\alpha & {\color{blue}[1]}\\ \beta \:=\:\arctan(y) &\Rightarrow& y \:=\:\tan\beta & {\color{blue}[2]}\end{array}$

The left side is: . $\arctan(x) + \arctan(y) \;=\;\alpha + \beta$

Take the tangent: . $\tan(\alpha + \beta) \;=\;\frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}$

Substitute ${\color{blue}[1]}$ and ${\color{blue}[2]}$: . $\tan(\alpha + \beta) \;=\;\frac{x+y}{1-xy}$

Hence: . $\alpha + \beta \;=\;\arctan\left(\frac{x+y}{1-xy}\right) + n\pi$

Therefore: . $\arctan(x) + \arctan(y) \;=\;\arctan\left(\frac{x+y}{1-xy}\right) + n\pi$

4. another small question guys
we have $+n\pi,n\in Z$ because $tan(x-\pi)=tan(x+\pi)=tan(x)$.
right thanks a lot.