<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] Use $\tan(A+B)=\tan(180^\circ-C)$:
$$\frac{(\tan A + \tan B)}{(1-\tan A \tan B)} = \frac{(\tan 180^\circ- \tan C)}{(1-\tan 180^\circ \tan C)}$$
Since $\tan 180^\circ=0$,
$$\frac{ (\tan A +\tan B)}{(1-\tan A\tan B) }= \frac{-\tan C}{1}$$
Therefore,
$$\tan A + \tan B = -\tan C + \tan A \tan B \tan C$$
Hence, the result
$$\tan A + \tan B + \tan C= \tan A \tan B \tan C$$
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