If ${\mathrm{x}}_1,{\mathrm{x}}_2,{\mathrm{x}}_3$ are the roots of ${\mathrm{x}}^3-6{\mathrm{x}}^2+11\mathrm{x}-6=0$ then ${\cot }^{-1}\left({\mathrm{x}}_1\right)+{\cot }^{-1}\left({\mathrm{x}}_2\right)+{\cot }^{-1}\left({\mathrm{x}}_3\right)$ is equal

$$\begin{gathered} {\mathrm{x}}^3-6{\mathrm{x}}^2+11\mathrm{x}-6=0 \mathrm{roots} \mathrm{are} 1,2,3 \\ {\mathrm{x}}_1=1 , {\mathrm{x}}_2=2 , {\mathrm{x}}_3=3 \\ ={\cot }^{-1}\left(1\right)+{\cot }^{-1}\left(2\right)+{\cot }^{-1}\left(3\right) \\ =\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.+{\tan }^{-1}\frac{1}{2}+{\tan }^{-1}\frac{1}{3}=\frac{\mathrm{\pi }}{4}+{\tan }^{-1}\left(\frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{1-{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}.{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\right) \\ =\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.+{\tan }^{-1}\left(1\right)=\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \end{gathered}$$