$\tan A + \tan B + \tan C = \tan A\tan B\tan C\,$ 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.

Here's another solution for the identity of Antonio Cagnoli :
We want to show that :
$\tan A + \tan B + \tan C = \tan A\times \tan B \times \tan C\ $  with $A+B+C=180^\circ=\pi$.
By definition we have : $\tan\alpha=\frac{\sin\alpha}{\cos\alpha}$ so here, we want to prove that :
$\frac{\sin A}{\cos A}+\frac{\sin B}{\cos B}+\frac{\sin C}{\cos C}=\frac{\sin A}{\cos A}\times\frac{\sin B}{\cos B}\times\frac{\sin C}{\cos C}$
$\Leftrightarrow$ $\frac{\sin A \cos B \cos C+ \sin B \cos A \cos C + \sin C \cos A \cos C}{\cos A \cos B \cos B}=\frac{\sin A \sin B \sin C}{\cos A \cos B \cos C}$
$\Leftrightarrow$ $\sin A \cos B \cos C+ \sin B \cos A \cos C + \sin C \cos A \cos B=\sin A \sin B \sin C$
However, 
$\sin A \cos B \cos C+ \sin B \cos A \cos C + \sin C \cos A \cos B=\cos C [\sin A \cos B + \sin B \cos A]+\sin C \cos A \cos B$
$\Leftrightarrow$
$\cos C [\sin A \cos B + \sin B \cos A]+\sin C \cos A \cos B=\cos C \sin(A+B)+\sin C \cos A \cos B$
$\Leftrightarrow$
$\cos C \sin(A+B)+\sin C \cos A \cos B=\cos C \sin(\pi - C) +\sin C \cos A \cos B$
$\Leftrightarrow$
$\cos C \sin(\pi - C) +\sin C \cos A \cos B=\cos C \sin C + \sin C \cos A \cos B$
$\Leftrightarrow$
$\cos C \sin C + \sin C \cos A \cos B=\sin C[\cos C + \cos A \cos B]$
$\Leftrightarrow$
$\sin C[\cos C + \cos A \cos B]=\sin C[\cos(\pi - (A+B)) + \cos A \cos B]$
$\Leftrightarrow$
$\sin C[\cos(\pi - (A+B)) + \cos A \cos B]=\sin C[-\cos(A+B) + \cos A \cos B]$
$\Leftrightarrow$
$\sin C[-\cos(A+B) + \cos A \cos B]=\sin C \sin A \sin B$.
We finally proved the equality !
Here https://play.google.com/books/reader?id=Lj51QYK1fIcC&printsec=frontcover&output=reader&hl=fr&pg=GBS.PA27 you can see the fastest answer !!!
