<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] HINT
$A+B+C = 180$
$A+B = 180 - C$ 
We'll apply tangent function: 
$\tan (A+B) = \tan (180 - C)$
We'll consider the identity:
$\tan(x+y) = \frac{\tan x + \tan y}{1-\tan x\tan y}$
$\frac{\tan A + \tan B}{1-\tan A\tan B} = \frac{\tan 180 - \tan C}{1+\tan 180\tan C}$
But $\tan 180 = 0$, therefore, we'll get:
$\frac{\tan A + \tan }{1-\tan A\tan B}$ = $\frac{0 - \tan C}{1+0}$
$\frac{\tan A + \tan B}{1-\tan A\tan B} = -\tan C$
We'll multiply by $(1-\tan A\tan B)$:
$\tan A + \tan B = -\tan C +\tan A\tan B\tan C$
Hence
$\tan A + \tan B+ \tan C = \tan A\tan B\tan C$
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