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Playing around on wolframalpha shows $\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3)=\pi$. I know $\tan^{-1}(1)=\pi/4$, but how could you compute that $\tan^{-1}(2)+\tan^{-1}(3)=\frac{3}{4}\pi$ to get this result?

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It is true in general that $\tan^{-1}(a) + \tan^{-1}(b) + \tan^{-1}(c) = \pi$ when $a+b+c=abc$ (and $a,b,c$ positive). This is the converse of what is proved here. – Henning Makholm Oct 19 '15 at 11:55
up vote 29 down vote accepted

$$\tan^{-1}(2)+\tan^{-1}(3)=\tan^{-1}{\left(\frac{2+3}{1-2\cdot 3}\right)}=\tan^{-1}(-1)=n\pi-\frac \pi 4,$$ where $n$ is any integer.

Now the principal value of $\tan^{-1}(x)$ lies in $[-\frac \pi 2, \frac \pi 2]$ precisely in $(0, \frac \pi 2)$ if finite $x>0$. So, the principal value of $\tan^{-1}(2)+\tan^{-1}(3)$ will lie in $(0, \pi) $.

So, the principal value of $\tan^{-1}(2)+\tan^{-1}(3)$ will be $\frac {3\pi} 4$.

Interestingly, the principal value of $\tan^{-1}(-1)$ is $-\frac {\pi} 4$.

But the general values of $\tan^{-1}(2)+\tan^{-1}(3)$ and $\tan^{-1}(-1)$ are same.

Alternatively, $$\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3)=\tan^{-1}{\left(\frac{1+2+3-1\cdot 2\cdot 3}{1-1\cdot 2- 2\cdot 3 -3\cdot 1}\right)}=\tan^{-1}(0)=m\pi$$, where $m$ is any integer.

Now the principal value of $\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3)$ will lie in $(0 ,\frac {3\pi} 2)$ which is $\pi$.

The principal value of $\tan^{-1}(0)$ is $0\neq \pi$.

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enter image description here

Consider $O=(0,0)$, $A=(1,1)$, $B=(-1,3)$, $D=(1,-3)$, $E=(1,0)$.

\begin{align} 2 &= \frac{AB}{AO} = \tan \angle AOB \\ 1 &= \frac{AE}{EO} = \tan \angle AOE \\ 3 &= \frac{DE}{DO} = \tan \angle DOE \end{align}

The points B, O and D are collinear, i.e. $\angle BOD = \tan^{-1}2+\tan^{-1}1+\tan^{-1}3 = \pi$.

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+1, I like showing facts in drawings - Somehow it makes facts easier to grasp. It may be nicer to explain the part: "The points B, O and E are collinear,..." a bit. Thanks. – NoChance Sep 16 '12 at 7:51
@EmmadKareem: Sorry, it should be B, O, D are collinear, and it is pretty obvious that they all fall on the same line $y = -3x$. – kennytm Sep 16 '12 at 8:06
Beautiful. Well done. – bubba Sep 16 '12 at 10:05
Can I ask how you made that picture? It looks really intuitive. :) – David Sep 16 '12 at 15:38
@David: GeoGebra. – kennytm Sep 16 '12 at 16:05

The simplest way is by using complex numbers. It is a trivial computation to show that $$(1+i)(1+2i)(1+3i)=-10$$ Now recall the geometric description of complex multiplication (multiply the lengths and add the angles), and take the argument on both sides of this equation. This gives $$\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3)=\pi$$

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Proof without word

$\tan^{-1} 1+\tan^{-1} 2+\tan^{-1} 3 =\pi$.

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Oh! I answered a duplicated question and here is the same nice solution (compare this). I don't know why this has only 4 upvotes :-| +1! – dtldarek Jan 7 '13 at 19:30
Agree with dtldarek, and trying to fix the upvote count. – Jyrki Lahtonen Jan 7 '13 at 21:36

Note that $$ \tan \left(\arctan(1+z) + \arctan\left(2 + z + z^2 \right) + \arctan \left( 3+3\,z+4\,{z}^{2}+2\,{z}^{3}+{z}^{4} \right) \right)=z $$ so that $$\arctan(1+z) + \arctan\left(2 + z + z^2 \right) + \arctan \left( 3+3\,z+4\,{z}^{2}+2\,{z}^{3}+{z}^{4} \right) = \arctan(z) + n \pi $$ for the appropriate integer $n$. For integers $z$ we get interesting arctan identities from this.

$$\eqalign{ \arctan(1) + \arctan\left(2\right)+ \arctan\left(3\right) &= \pi \cr \arctan(2) + \arctan(4) + \arctan(13) &= \arctan(1) + \pi \cr \arctan(3) + \arctan(8) + \arctan(57) &= \arctan(2) + \pi \cr \arctan(4) + \arctan(14) + \arctan(183) &= \arctan(3) + \pi \cr}$$ etc.

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How did you manage the first equation? – Nastassja Sep 16 '12 at 7:09
I asked Maple for integer solutions of $$\frac{a+b+c-abc}{1-ab-bc-ca} = d$$ – Robert Israel Sep 16 '12 at 7:24
Thanks! ${}{}{}$ – Nastassja Sep 16 '12 at 7:25
I know that this an old answer, but would you please explain how the integer solutions of $$\frac{a+b+c-abc}{1-ab-bc-ca} = d$$ relate to your first identity? I can't see it on my own :( – math.n00b Jul 23 '14 at 9:58
Maple gave me a parametric solution which simplifies to $$\left\{ a={{\it \_Z1}}^{4}+2\,{{\it \_Z1}}^{3}+4\,{{\it \_Z1}}^{2}+3 \,{\it \_Z1}+3,b={{\it \_Z1}}^{2}+{\it \_Z1}+2,c=1+{\it \_Z1},d={\it \_Z1} \right\} $$ – Robert Israel Jul 23 '14 at 15:12

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