# $\tan A + \tan B + \tan C = \tan A\tan B\tan C\,$ 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.

• There is another geometric proof here: maa.org/programs/faculty-and-departments/… May 17, 2015 at 18:58
• You are so close. Write it as $-\tan(C) = \frac{\tan A+\tan B}{1-\tan A\tan B}$ and then multiply by $1-\tan A\tan B$. May 17, 2015 at 19:34

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

• I wonder whether the nice symmetric algebra here encodes a similarly nice geometric intuition ... May 17, 2015 at 14:56
• sir,can you pls guide me on how we can prove the converse , ie. if $tan(a)+tan(b)+tan(c)=tan(a)tan(b)tan(c)$ then there exist a triangle ...i am totally confused with the converse part and i have no idea on how to prove t ,please help me Jul 25, 2020 at 5:41
• The converse is not necessarily true. For example: $a=b=c=0$ or $a=0$ and $b=2\pi-c$.
– robjohn
Jul 25, 2020 at 12:03
• @hmakholmleftoverMonica this comment made my day. Absolutely flawless. Dec 27, 2022 at 20:48

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

$A+B=180-C$

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

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

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

\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}

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 !

• 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. Jul 12, 2018 at 7:45
• @ChrisSteinbeckBell Could you be more precise with the last formula ? Jul 12, 2018 at 12:16
• 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. Jul 12, 2018 at 20:15
• @ChrisSteinbeckBell Ok I see. Maybe it's Cagnoli again. I suggest you to read the google books in my answer ! Jul 13, 2018 at 5:56
• 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. Jul 16, 2018 at 9:01

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.

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