I have two angles $A_1, A_2 > 0$, and $A_1+A_2 < \pi$, is it possible to find an $A_0$ such that
$$\tan(A_0),\ \tan(A_0+A_1),\ \tan(A_0+A_1+A_2)$$
forms an arithmetic sequence on the same continuous range of tangent?
I have been looking for a formula for sum of tangents (not tangent of sum), but I have not been successful so far.
(Resolving my own question)
Let the first term $\tan(A_0) = a$, term difference be $d$, then
$$\begin{align*} \cot A_1 &= \cot(A_0+A_1-A_0)\\ &= \frac{1+\tan(A_0+A_1)\tan A_0}{\tan(A_0+A_1)-\tan A_0}\\ &= \frac{1+(a+d)a}{d}\\ \cot A_2 &= \cot(A_0+A_1+A_2-A_0-A_1)\\ &= \frac{1+\tan(A_0+A_1+A_2)\tan(A_0+A_1)}{\tan(A_0+A_1+A_2)-\tan(A_0+A_1)}\\ &= \frac{1+(a+2d)(a+d)}{d}\\ \end{align*}$$
then $$\begin{align*} \cot A_2 - \cot A_1 &= \frac{1+(a+2d)(a+d)}d - \frac{1+(a+d)a}d\\ &= \frac{2d(a+d)}d\\ &= 2(a+d)\\ &= 2\tan(A_0+A_1) \end{align*}$$
A value of $A_0$ can be taken as $$A_0 = \arctan \left(\frac{\cot A_2 - \cot A_1}2\right)-A_1$$