Question

Show geometrically this is true:

Answer 1

\[\Large \arctan(1)+\arctan(2)+\arctan(3)=\pi\]

Answer 2

Please note that if I make the tangent of both sides, I get:
\[\begin{gathered}
\tan \left[ {\left( {\arctan (1) + \arctan (2)} \right) + \arctan (3)} \right] = \tan \left( \pi \right) \hfill \\
\frac{{\tan \left( {\arctan (1) + \arctan (2)} \right) + 3}}{{1 - \tan \left( {\left( {\arctan (1) + \arctan (2)} \right)} \right) \cdot 3}} = \tan \left( \pi \right) \hfill \\
\frac{{\frac{{1 + 2}}{{1 - 1 \cdot 2}} + 3}}{{1 - \frac{{1 + 2}}{{1 - 1 \cdot 2}} \cdot 3}} = \tan \left( \pi \right) \hfill \\
0 = 0 \hfill \\
\end{gathered} \]

Answer 3

its gotta be geometrically brooo