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.

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.
