Why is $\arctan\frac{x+y}{1-xy} = \arctan x +\arctan y$? Why is $\arctan\frac{x+y}{1-xy} = \arctan x +\arctan y$?
It is said that this is derived from trigonometry, but I couldn't find why this is the case.
 A: The following identity is useful for your purpose 

$$ \tan(\alpha \pm \beta) = \frac{\tan \alpha \pm \tan \beta}{1 \mp \tan \alpha \tan \beta}. $$

To prove it just use the identity
$ \tan(t) = \frac{\sin(t)}{\cos(t)}$ and the identities for $\sin(a\pm b)$ and $\cos(a\pm b)$.
A: Computing with complex numbers:
$$
(1+xi)(1+yi)=(1-xy)+(x+y)i
$$
Take arg on both sides
$$
\arctan x + \arctan y
= \arg(1+xi)+\arg(1+yi)
= \arg((1+xi)(1+yi))
\\
= \arg((1-xy)+(x+y)i)
= \arctan\frac{x+y}{1-xy}
$$
A: Here's a proof that has essentially nothing to do with trigonometry:
Hold $y$ constant and differentiate the function $$f(x) = \arctan{\frac{x + y}{1 - xy}} - \arctan{x} - \arctan{y}$$ to find that
\begin{align}f'(x) &= \frac{1}{1 + \left(\frac{x + y}{1 - xy}\right)^2} \cdot \left(\frac{(1 - xy) - (x + y)(-y)}{(1 - xy)^2}\right) - \frac{1}{1 + x^2} \\
&= \frac{(1 - xy)^2}{(1 - xy)^2 + (x + y)^2} \cdot \left(\frac{1 - xy + xy + y^2}{(1 - xy)^2}\right) - \frac{1}{1 + x^2} \\
&= \frac{1 + y^2}{1 - 2xy + x^2y^2 + x^2 + 2xy + y^2} - \frac{1}{1 + x^2} \\
&= \frac{1 + y^2}{1 + x^2 + y^2 + x^2y^2} - \frac{1}{1 + x^2} \\
&= \frac{1 + y^2}{(1 + y^2) + x^2 (1 + y^2)} - \frac{1}{1 + x^2} \\
&= \frac{1}{1 + x^2} - \frac{1}{1 + x^2} = 0
\end{align}
So $f$ is a constant. Letting $x = 0$, we find that $f(0) = 0$ and the identity follows.
A: Do NOT blindly use this for calculations. Example
$\arctan(5000)+\arctan(5000) \sim \pi/2+\pi/2=\pi$
with the formula  quoted you get
$\arctan(10000/(1-25\times10^6))\sim\arctan(-1/2500)\sim~0$. Like Andre Nicolas said, you need to considr arctan ranges
A: Consider $$A=\arctan(x)+\arctan(y),$$ then $$\tan(A)=\tan(\arctan(x)+\arctan(y)),$$
which is also equal to:$$\frac{\tan(\arctan(x))+\tan(\arctan(x))}{1-\tan(\arctan(x)) + \tan(\arctan(y)},$$
which simplifies to giv:$$\tan(A)=\frac{x+y}{1-xy}.$$
Thus,$$\arctan(x)+\arctan(y)=A=\arctan(\tan(A))=\arctan\frac{x+y}{1-xy}.$$
A: Ok fine, mlc. I'll answer the question. Here is a trigonometric proof. No calculus required.
Statement: $\arctan(\frac {x+y} {1-xy})=\arctan(x)+\arctan(y)$ whenever $|xy|<1$.
Proof. Choose $(x,y)\in\big\{|xy|<1\big\}$ arbitrarily. This formula is obvious if $x=0,$ so first consider when $x>0$. We prove that $\arctan(x)+\arctan(y)\in\Big(\frac{-\pi}{2},\frac{\pi}{2}\Big)$. Note $|xy|<1$ if and only if $-\frac{1}{x}<y<\frac1x$. Since $\arctan$ is an increasing and odd function, it follows $$-\arctan\Big(\frac{1}{x}\Big)<\arctan(y)<\arctan\Big(\frac{1}{x}\Big).$$ Indeed, $\arctan(x)+\arctan(\frac{1}{x})=\frac{\pi}{2},$ and so $$-\frac\pi2<2\arctan(x)-\frac\pi2<\arctan(x)+\arctan(y)<\frac\pi2.$$ This shows $\arctan(x)+\arctan(y)$ belongs to the range of $\arctan$, $\Big(\frac{-\pi}{2},\frac{\pi}{2}\Big)$. The angle sum identity tells us $$\begin{equation}\begin{split}\tan(\arctan(x)+\arctan(y)) &=\frac{\tan(\arctan(x))+\tan(\arctan(y))}{1-\tan(\arctan(x))\tan(\arctan(y))} \\
& =\frac{x+y}{1-xy}\end{split}\end{equation}$$
Hence $$\begin{equation}\begin{split}\arctan(x)+\arctan(y)&=\arctan(\tan(\arctan(x)+\arctan(y))) \\ &=\arctan\Big(\frac{x+y}{1-xy}\Big)\end{split}\end{equation}$$ This proves the identity for $x>0,$ so now assume $x<0$. Then $-x>0$ and $|xy|<1$ $\iff$ $-\frac{1}{x}<-y<\frac{1}{x}.$ Finally, $$\begin{equation}\begin{split}\arctan(x)+\arctan(y) & =-\big(\arctan(-x)+\arctan(-y)\big) \\ & =-\arctan\Big(\frac{-x-y}{1-(-x)(-y)}\Big) \\ & =\arctan\Big(\frac{x+y}{1-xy}\Big)\end{split}\end{equation}$$ The result follows.
