<s>[INST] 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 
$$ \frac{ \tan\left(x\right)+ \tan\left(y\right)}{1- \tan(x) \tan(y)} = \tan\left(x+y \right). 
$$
 [/INST] 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 $C(y)=\arctan{y}$.
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