Arithmetic sequence of tangent values 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
$$A_0 = \arctan \left(\frac{\cot A_2 - \cot A_1}2\right)-A_1$$
