If $\triangle{ABC}$ is right, without loss of generality, let's assume $A=90^\circ$. Then $\cot A = 0$ and $\cot B=\tan C$ because $B+C=90^\circ$. Then $$\cot A\cot B +\cot B\cot C + \cot C\cot A = 0+\cot B \cot C + 0 =1$$ Otherwise, if none of the three angles is right, then all of $\tan A$, $\tan B$ and $\tan C$ are well defined. Multiplying both sides of the to-be-claimed relations by $\tan A\tan B\tan C$ yields $$\tan A + \tan B + \tan C = \tan A\tan B\tan C$$ This is known to hold by %%HREF%%3896%%. Therefore, we conclude the relation always holds.