<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] For any angle $\theta$, let $c_\theta = \cos\theta, s_\theta = \sin\theta$ and $t_\theta = \tan\theta$, we have:
$$\begin{align} 
e^{iA} e^{iB} e^{iC}
& = ( c_A + i s_A )(c_B + i s_B)(c_C + is_C)\\[6pt]
& = c_A c_B c_C (1 + i t_A)(1 + i t_B )(1 + i t_C)\\
& = c_A c_B c_C \bigg[ \big( 1 - (t_A t_B + t_B t_C + t_C t_A ) \big) + i \big( t_A + t_B + t_C - t_A t_B t_C \big)\bigg]
\end{align}$$
This implies 
$$\frac{\Im(e^{iA} e^{iB} e^{iC})}{\Re(e^{iA} e^{iB} e^{iC})} = \frac{t_A + t_B + t_C - t_A t_B t_C}{1 - t_A t_B - t_B t_C - t_C t_A}\tag{*}$$
One the other hand, 
$$e^{iA} e^{iB} e^{iC} = e^{i(A+B+C)} = c_{A+B+C}(1 + i t_{A+B+C}),$$
The L.H.S of $(*)$ is simply $t_{A+B+C}$. From this, we get the addition formula of tangent
for three angles:
$$t_{A+B+C} = \frac{t_A + t_B + t_C - t_A t_B t_C}{1 - t_A t_B - t_B t_C - t_C t_A}\\
\iff\tan(A+B+C) = \frac{\tan A + \tan B + \tan C - \tan A\tan B\tan C}{1 - \tan A \tan B - \tan B\tan C - \tan C\tan A}
$$
In particular, this means
$$\tan A + \tan B + \tan C = \tan A\tan B\tan C \iff \tan(A+B+C) = 0$$
If we have further information that $0 < A+B+C < 360^{\circ}$, then this equivalence
can be rewritten as:
$$\tan A + \tan B + \tan C = \tan A\tan B\tan C \iff A+B+C = 180^\circ$$
</s>