<s>[INST] $\tan A + \tan B + \tan C = \tan A\tan B\tan C\,$ 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.
 [/INST] \begin{eqnarray}
\tan A+\tan B+\tan C&=&\frac{\sin A\cos B+\sin B\cos A}{\cos A\cos B}+\tan(180^\circ-A-B)\\
&=&\frac{\sin(A+B)}{\cos A\cos B}-\frac{\sin(A+B)}{\cos(A+B)}\\
&=&\sin(A+B)\frac{\cos(A+B)-\cos A\cos B}{\cos A\cos B\cos(A+B)}\\
&=&-\frac{\sin A\sin B\sin(A+B)}{\cos A\cos B\cos(A+B)}\\
&=&-\tan A\tan B\tan(A+B)\\
&=&\tan A\tan B\tan C.
\end{eqnarray}
</s>