$\tan A + \tan B + \tan C = \tan A\tan B\tan C\,$ in a triangle I want to prove this (where each angle may be negative or greater than $180^\circ$):

When $A+B+C = 180^\circ$ \begin{equation*} \tan A  + \tan B + \tan C = \tan A\:\tan B\:\tan C. \end{equation*}

We know that
\begin{equation*}\tan(A+B) = \frac{\tan A+\tan B}{1-\tan A\tan B}\end{equation*} and that \begin{equation*}\text{and that}~A+B = 180^\circ-C.\end{equation*}
Therefore $\tan(A+B) = -\tan C.$
From here, I got stuck.

$$A+B=180^\circ-C\\
\tan(A+B)=\tan(180^\circ-C)\\
\frac{\tan A+\tan B}{1-\tan A\tan B}  =-\tan C\\
\tan A +\tan B +\tan C =\tan A \tan B \tan C.$$

