$\tan A + \tan B + \tan C = \tan A\tan B\tan C\,$ in a triangle

Note that $$ \mathrm{Im}\left(e^{i\pi}\right)=0\tag{1} $$ Thus, if $a+b+c=\pi$, $$ \begin{align} 0 &=\mathrm{Im}\left(e^{ia}e^{ib}e^{ic}\right)\\ &=\mathrm{Im}\Big(\big(\cos(a)+i\sin(a)\big)\big(\cos(b)+i\sin(b)\big)\big(\cos(c)+i\sin(c)\big)\Big)\\[4pt] &=\sin(a)\cos(b)\cos(c)+\cos(a)\sin(b)\cos(c)+\cos(a)\cos(b)\sin(c)\\ &-\sin(a)\sin(b)\sin(c)\tag{2} \end{align} $$ Dividing $(2)$ by $\cos(a)\cos(b)\cos(c)$ yields $$ \tan(a)+\tan(b)+\tan(c)=\tan(a)\tan(b)\tan(c)\tag{3} $$

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

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

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.

HINT:

Using $\displaystyle \tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B},$

we can prove $$\tan(A+B+C)=\frac{\sum_\text{cyc}\tan A-\prod \tan A}{1-\sum_\text{cyc}\tan A\tan B}$$

Now, if $A+B+C=n180^\circ,$ where $n$ is any integer we know $\tan(n180^\circ)=0$

For any angle $\theta$, let $c_\theta = \cos\theta, s_\theta = \sin\theta$ and $t_\theta = \tan\theta$, we have:

$$\begin{align} e^{iA} e^{iB} e^{iC} & = ( c_A + i s_A )(c_B + i s_B)(c_C + is_C)\\[6pt] & = c_A c_B c_C (1 + i t_A)(1 + i t_B )(1 + i t_C)\\ & = c_A c_B c_C \bigg[ \big( 1 - (t_A t_B + t_B t_C + t_C t_A ) \big) + i \big( t_A + t_B + t_C - t_A t_B t_C \big)\bigg] \end{align}$$ This implies $$\frac{\Im(e^{iA} e^{iB} e^{iC})}{\Re(e^{iA} e^{iB} e^{iC})} = \frac{t_A + t_B + t_C - t_A t_B t_C}{1 - t_A t_B - t_B t_C - t_C t_A}\tag{*}$$

One the other hand,

$$e^{iA} e^{iB} e^{iC} = e^{i(A+B+C)} = c_{A+B+C}(1 + i t_{A+B+C}),$$ The L.H.S of $(*)$ is simply $t_{A+B+C}$. From this, we get the addition formula of tangent for three angles:

$$t_{A+B+C} = \frac{t_A + t_B + t_C - t_A t_B t_C}{1 - t_A t_B - t_B t_C - t_C t_A}\\ \iff\tan(A+B+C) = \frac{\tan A + \tan B + \tan C - \tan A\tan B\tan C}{1 - \tan A \tan B - \tan B\tan C - \tan C\tan A} $$ In particular, this means $$\tan A + \tan B + \tan C = \tan A\tan B\tan C \iff \tan(A+B+C) = 0$$

If we have further information that $0 < A+B+C < 360^{\circ}$, then this equivalence can be rewritten as: $$\tan A + \tan B + \tan C = \tan A\tan B\tan C \iff A+B+C = 180^\circ$$