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?

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
