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I want to prove

\begin{equation*} \tan A + \tan B + \tan C = \tan A\tan B\tan C \quad\text{when } A+B+C = 180^\circ \end{equation*}

We know that

\begin{equation*} \tan(A+B) = \frac{\tan A+\tan B}{1-\tan A\tan B}~\text{and that}~A+B = 180^\circ-C. \end{equation*}

Therefore $\tan(A+B) = -\tan C.$ From here, I get stuck. Please help.

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  • 2
    $\begingroup$ There is another geometric proof here: maa.org/programs/faculty-and-departments/… $\endgroup$ – Steve Kass May 17 '15 at 18:58
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    $\begingroup$ You are so close. Write it as $-\tan(C) = \frac{\tan A+\tan B}{1-\tan A\tan B}$ and then multiply by $1-\tan A\tan B$. $\endgroup$ – marty cohen May 17 '15 at 19:34

10 Answers 10

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

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    $\begingroup$ I wonder whether the nice symmetric algebra here encodes a similarly nice geometric intuition ... $\endgroup$ – Henning Makholm May 17 '15 at 14:56
  • $\begingroup$ A nice alternative $\endgroup$ – Simon S Oct 10 '15 at 12:33
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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

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

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Here is a geometric proof, for the case that all three angles are acute:

diagram

$QRUV$ 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.

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

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

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$A+B=180-C$

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

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

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\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}

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

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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 !!!

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  • $\begingroup$ I didn't know this identity had a name. I searched this wiki entry but the article is too short and doesn't mention anything. But this other entry shows a list of his published works. Yet there is no specific mention of the name neither the identity. Do you perhaps have any sources where does the name comes from?. There is another identity related with the sum of the double angles of sines, to which is also well known, does this also belong to him?. I'm very curious. $\endgroup$ – Chris Steinbeck Bell Jul 12 '18 at 7:45
  • $\begingroup$ @ChrisSteinbeckBell Could you be more precise with the last formula ? $\endgroup$ – Maman Jul 12 '18 at 12:16
  • $\begingroup$ I was referring to $\sin 2\omega + \sin 2\phi + \sin 2\psi = 4 \sin \omega \sin \phi \sin \psi$. So is this related to Antonio Cagnoli as well?. Is it now more clear?. I like to know about the history. $\endgroup$ – Chris Steinbeck Bell Jul 12 '18 at 20:15
  • $\begingroup$ @ChrisSteinbeckBell Ok I see. Maybe it's Cagnoli again. I suggest you to read the google books in my answer ! $\endgroup$ – Maman Jul 13 '18 at 5:56
  • $\begingroup$ I've checked the book your mentioned, it is in french though (I don´t know french). Anyway I opened a question regarding this in History of Science and Math stack if you're interested I'm yet curious if you know other sources than that book. $\endgroup$ – Chris Steinbeck Bell Jul 16 '18 at 9:01
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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.

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