$\tan A + \tan B + \tan C = \tan A\tan B\tan C\,$ 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.
 A: 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 $\tan(n180^\circ)=0$
A: 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
$$
\tan(a)+\tan(b)+\tan(c)=\tan(a)\tan(b)\tan(c)\tag{3}
$$
A: 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:
$$\tan A + \tan B + \tan C = \tan A\tan B\tan C \iff A+B+C = 180^\circ$$
A: Use $\tan(A+B)=\tan(180^\circ-C)$:
$$\frac{(\tan A + \tan B)}{(1-\tan A \tan B)} = \frac{(\tan 180^\circ- \tan C)}{(1-\tan 180^\circ \tan C)}$$
Since $\tan 180^\circ=0$,
$$\frac{ (\tan A +\tan B)}{(1-\tan A\tan B) }= \frac{-\tan C}{1}$$
Therefore,
$$\tan A + \tan B = -\tan C + \tan A \tan B \tan C$$
Hence, the result
$$\tan A + \tan B + \tan C= \tan A \tan B \tan C$$
A: $A+B=180-C$
$\tan(A+B)=\tan(180-C)$
$[\tan(A)+\tan(B)]/[1-\tan(A)\tan(B)]=-\tan(C)$
 $\tan(A)+\tan(B)=-\tan(C)+\tan(A)\tan(B)\tan(C)$
 $\tan(A)+\tan(B)+\tan(C)=\tan(A)\tan(B)\tan(C)$
A: 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
$\tan A + \tan B+ \tan C = \tan A\tan B\tan C$
A: 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.
A: \begin{eqnarray}
\tan A+\tan B+\tan C&=&\frac{\sin A\cos B+\sin B\cos A}{\cos A\cos B}+\tan(180^\circ-A-B)\\
&=&\frac{\sin(A+B)}{\cos A\cos B}-\frac{\sin(A+B)}{\cos(A+B)}\\
&=&\sin(A+B)\frac{\cos(A+B)-\cos A\cos B}{\cos A\cos B\cos(A+B)}\\
&=&-\frac{\sin A\sin B\sin(A+B)}{\cos A\cos B\cos(A+B)}\\
&=&-\tan A\tan B\tan(A+B)\\
&=&\tan A\tan B\tan C.
\end{eqnarray}
A: 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 !!!
A: 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.
A: $$A+B=180^\circ-C\\
\tan(A+B)=\tan(180^\circ-C)\\
\frac{\tan A+\tan B}{1-\tan A\tan B}  =-\tan C\\
\tan A +\tan B +\tan C =\tan A \tan B \tan C.$$

