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

Maybe a little bit obvious but what I would do is the following:
$\textrm{Let:}$ 
$$\omega + \phi + \psi = \pi$$
Then take the tangent function to both sides.
$$\tan \left ( \omega + \phi + \psi \right) = \tan \left ( \pi \right)$$
Since $\tan \pi = 0$, then:
$$\tan \left ( \omega + \phi + \psi \right) = \tan \left ( \pi \right)$$
Now group:
$$\tan \left ( \left ( \omega + \phi + \right) + \psi \right) = \tan \left ( \pi \right)$$
Resolving we have:
$$\frac{\tan \left (\omega + \phi \right) + \tan \left (\psi \right) }{1-\tan \left( \omega + \phi \right) \tan \left (\psi \right)} = 0$$
In order to make the whole equation to zero, the numerator has to be zero as well, therefore just replace:
$$ \tan \left (\omega + \phi \right) + \tan \left (\psi \right) = 0$$
By expanding it:
$$\tan \left (\omega + \phi \right) = - \tan \left (\psi \right )$$
$$\frac{\tan \omega + \tan \phi}{1- \tan \omega \tan \phi} = - \tan \psi$$
$$\tan \omega + \tan \phi = \left( 1- \tan \omega \tan \phi \right) \left ( - \tan \psi \right )$$
Don't worry, we're almost there:
$$\tan \omega + \tan \phi = - \tan \psi + \tan \omega \tan \phi \tan \psi$$
$$\tan \omega + \tan \phi + \tan \psi = \tan \omega \tan \phi \tan \psi$$
Therefore we have proved the identity!.
By the way I used the angles $\omega$, $\phi$ and $\psi$ as I feel more comfortable working with them but in your case you may want them to be replaced by the letters $\textrm{A, B and C}$.
I hope this have helped you.
