In ${∆}^{\mathrm{le}} \mathrm{ABC}$ if $\mathrm{CosA} \mathrm{CosB} \mathrm{CosC} = 1/3$ then $\mathrm{TanA} \mathrm{TanB} + \mathrm{TanB} \mathrm{TanC} + \mathrm{TanC} \mathrm{TanA}=$

$$\begin{gathered} \begin{array}{lcl}\mathrm{A}+\mathrm{B}+\mathrm{C}\begin{array}{cl}=& \mathrm{\pi }\end{array}& & \\ \frac{\cos (\mathrm{A}+\mathrm{B}+\mathrm{C})}{\cos \mathrm{A} \cos \mathrm{B} \cos \mathrm{C}}& & \\ \begin{array}{cl}=& 1-\tan \mathrm{A} \tan \mathrm{B}-\tan \mathrm{B} \tan \mathrm{C}-\tan \mathrm{C} \tan \mathrm{A}\end{array}& & \\ \tan \mathrm{A} \tan \mathrm{B}+\tan \mathrm{B} \tan \mathrm{C}+\tan \mathrm{C} \tan A& & \\ \begin{array}{cl}=& 1-\frac{\cos \mathrm{\pi }}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\end{array} \\ \begin{array}{cl}=& 1+3=4\end{array}& & \end{array} \end{gathered}$$