\(0\leq\Arctan\frac{1}{2}+\Arctan\frac{1}{5}<\Arctan1+\Arctan1=\frac{\pi}{2}\) et

\[\tan\left(\Arctan\frac{1}{2}+\Arctan\frac{1}{5}\right)=\frac{\frac{1}{2}+\frac{1}{5}}{1-\frac{1}{2}\times\frac{1}{5}} =\frac{7}{9}.\] Comme \(\Arctan\frac{1}{2}+\Arctan\frac{1}{5}\in[0,\frac{\pi}{2}[\), on a donc \(\Arctan\frac{1}{2}+\Arctan\frac{1}{5}=\Arctan\frac{7}{9}\). De même, \(\Arctan\frac{7}{9}+\Arctan\frac{1}{8}\in[0,\frac{\pi}{2}[\) et

\[\tan\left(\Arctan\frac{7}{9}+\Arctan\frac{1}{8}\right)=\frac{\frac{7}{9}+\frac{1}{8}}{1-\frac{7}{9}\times\frac{1}{8}}=\frac{65}{65} =1,\] et donc \(\Arctan\frac{7}{9}+\Arctan\frac{1}{8}=\Arctan1=\frac{\pi}{4}\). Finalement,

\(\Arctan\frac{1}{2}+\Arctan\frac{1}{5}+\Arctan\frac{1}{8}=\frac{\pi}{4}.\)

exo7 5089