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

Question

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.

Answer 1

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 $$ \tan(a)+\tan(b)+\tan(c)=\tan(a)\tan(b)\tan(c)\tag{3} $$

Answer 2

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

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

Answer 3

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.

Answer 4

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 $\tan(n180^\circ)=0$

Answer 5

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: $$\tan A + \tan B + \tan C = \tan A\tan B\tan C \iff A+B+C = 180^\circ$$

Answer 6

$A+B=180-C$

$\tan(A+B)=\tan(180-C)$

$[\tan(A)+\tan(B)]/[1-\tan(A)\tan(B)]=-\tan(C)$ $\tan(A)+\tan(B)=-\tan(C)+\tan(A)\tan(B)\tan(C)$ $\tan(A)+\tan(B)+\tan(C)=\tan(A)\tan(B)\tan(C)$

Answer 7

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

Answer 8

\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}

Answer 9

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 !!!

Answer 10

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.

Answer 11

$$A+B=180^\circ-C\\ \tan(A+B)=\tan(180^\circ-C)\\ \frac{\tan A+\tan B}{1-\tan A\tan B} =-\tan C\\ \tan A +\tan B +\tan C =\tan A \tan B \tan C.$$