Additivity of $\arctan(\frac{x+y}{1-xy})$ Show in two different way that,
$$\arctan\left({\frac{x+y}{1-xy}}\right) = \arctan\left(x\right) + \arctan\left(y\right)
$$
The first way I know derive from the addition formula of tangent 
$$ \frac{ \tan\left(x\right)+ \tan\left(y\right)}{1- \tan(x) \tan(y)} = \tan\left(x+y \right). 
$$
 A: Differentiating,
$$ \begin{align}
\frac{d}{dx} \arctan{\left( \frac{x+y}{1-xy} \right)} &= \frac{1}{1+(x+y)^2/(1-xy)^2} \left( \frac{1}{1-xy} - \frac{-(x+y)y}{(1-xy)^2} \right) \\
&= \frac{ 1-xy + xy+y^2 }{(1-xy)^2 + (x+y)^2} \\
&= \frac{ 1+y^2 }{1-2xy+x^2y^2+x^2+2xy+y^2} \\
&= \frac{ 1+y^2 }{1+x^2+y^2+x^2 y^2} \\
&= \frac{ 1+y^2 }{(1+y^2)(1+x^2)} = \frac{1}{1+x^2},
\end{align} $$
and then integrating gives
$$ \arctan{\left( \frac{x+y}{1-xy} \right)} = \arctan{x}+C(y). $$
Putting $x=0$ gives $C(y)=\arctan{y}$.
A: There's always a certain "crystal ball" aspect to a question such as this, but another approach is to show that for each (fixed) real number $y$, the function
$$
f(x) = \arctan\left(\frac{x + y}{1 - xy}\right) - \arctan x - \arctan y
$$
has vanishing derivative (throughout its domain), and vanishes at some point (such as where $y = -x$).
(Strictly speaking, of course, the two are not equal, since the left-hand side is undefined where $xy = 1$ and the right-hand side is defined for all real $x$ and $y$.)
A: $$(1+ix)(1+iy)=(1-xy)+i(x+y)$$
$$\arg(z_1z_2)=\arg(z_1)+\arg(z_2)$$
$$\arg(a+bi)=\arctan \frac{b}{a}$$
$$\implies\arctan x + \arctan y = \arctan \frac{x+y}{1-xy}$$
Remarks:


*

*Depending on your perspective, this might be more of a proof of the 2nd line.

*The third line is not entirely true, as it can require some adjustment depending on the signs of $a,b$

*This is also true of the final result, actually.

