Why does $\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3)=\pi$? 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?
 A:  
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. $\angle BOD = \tan^{-1}2+\tan^{-1}1+\tan^{-1}3 = \pi$.
A: Proof without word
$\tan^{-1} 1+\tan^{-1} 2+\tan^{-1} 3 =\pi$.

A: 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 
$$\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3)=\pi$$
A: $$\tan^{-1}(2)+\tan^{-1}(3)=\tan^{-1}{\left(\frac{2+3}{1-2\cdot 3}\right)}=\tan^{-1}(-1)=n\pi-\frac \pi 4,$$ 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 $0\neq \pi$.
A: 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.
A: Consider, $z_1=   \frac{1+2i}{\sqrt{5}}$, $z_2= \frac{1+3i}{\sqrt{10} }$, and $z_3= \frac{1+i}{\sqrt{2} }$, then:
$$ z_1 z_2 z_3 = \frac{1}{10} (1+2i)(1+3i)(1+i)=-1 $$
Take arg of both sides and use property that $\arg(z_1 z_2 z_3) = \arg(z_1) + \arg(z_2)  + \arg(z_3)$:
$$ \arg(z_1) + \arg(z_2) + \arg(z_3) = -1$$
The LHS we can write as:
$$ \tan^{-1} ( \frac{2}{1})  +\tan^{-1} ( \frac{3}{1} ) + \tan^{-1} (1)  = \pi$$
Tl;dr: Complex number multiplication corresponds to tangent angle addition
