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?

Answer

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

Attribution
Source : Link , Question Author : Clayton Kershaw , Answer Author : kennytm

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