<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] Here is a geometric proof, for the case that all three angles are acute:

$QRUV$ are collinear because $B+90^\circ+(90^\circ-B)=180^\circ$.
$STV$ are collinear because $A+B+C=180^\circ$, so $\angle QSV=\angle UTV=C$.
Similar triangles $\triangle PQR\sim\triangle TRS$ and $\triangle RTU \sim \triangle SRQ$ give $\displaystyle
\frac{QP}{RQ} = \frac{RT}{SR} = \frac{TU}{RQ}$, and therefore $TU=QP=1$.
Then,
$$\begin{align}& \tan A + \tan B + \tan C = QR+RU+UV = QV \\
&= QP \frac{QR}{QP}\, \frac{QS}{QR} \, \frac{QV}{QS} = 1 \cdot \tan(A) \tan(B) \tan(C) \end{align}$$

When one of the angles is obtuse, let it (without loss of generality) be $C$. Then a similar diagram can be drawn, except that $V$ is to the left of $Q$, and $UV$, $QV$ count as negative lengths.
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