Is there any formula for $\operatorname{arctan}{(a+b)}$

Question

Is there any formula for $$\operatorname{arctan}{(a+b)}$$ I know couple of formulas for trigonometric functions. But I don't know if such formulas exists for inverse trigonometric functions. I don't have a clue where to start with. Any hint will be appreciated.

Answer 1

The following formula is well known and you can find it here. $$\arctan(a) + \arctan(b) = \arctan\left(\frac{a+b}{1-ab}\right)$$ (of course up to a multiple of $\pi$ with $ab\neq 1$). Then $$\arctan(a+b) + \arctan(a-b) = \arctan\left(\frac{2 a}{1-a^2+b^2}\right)$$

$$\arctan(a+b) - \arctan(a-b) = \arctan\left(\frac{2 b}{1+a^2-b^2}\right)$$

Hence we have

\begin{align*} 2 \arctan(a+b) &= \arctan\left(\frac{2 a}{1-a^2+b^2}\right)+\arctan\left(\frac{2 b}{1+a^2-b^2}\right)\\[2mm] \arctan(a+b) &= \frac{1}{2}\arctan\left(\frac{2 a}{1-a^2+b^2}\right)+\frac{1}{2}\arctan\left(\frac{2 b}{1+a^2-b^2}\right) \end{align*}