<s>[INST] 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?
 [/INST] 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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