Problem - 3896
If $\triangle{ABC}$ is not a right triangle, show
\begin{equation}
\tan A + \tan B + \tan C =\tan A \tan B \tan C
\end{equation}
Given $\triangle{ABC}$ is not right, $\tan A$, $\tan B$, and $\tan C$ are all well-defined. Then,
\begin{align*}
&\tan A + \tan B + \tan C\\
=\quad & \tan(A+B)(1-\tan A\tan B) + \tan C\\
=\quad & \tan (180^\circ - C) (1-\tan A \tan B) + \tan C\\
=\quad & -\tan C ( 1- \tan A \tan B) + \tan C\\
=\quad &\tan A \tan B\tan C
\end{align*}