In $\triangle ABC$, if $\cos A\cos B\cos C=\frac{1}{3}$, then $\tan A\tan B+\tan B \tan C+\tan C\tan A =\text{???}$ 
In $\triangle ABC$, if 
  $$\cos A \cos B \cos C=\frac{1}{3}$$ 
  then can we find value of 
  $$\tan A\tan B+\tan B \tan C+\tan C\tan A\ ?$$

Please give some hint. I am not sure if $\tan A \tan B+\tan B \tan C+\tan C \tan A$ will be constant under given condition. 

Let $$S=\tan A\tan B+\tan B\tan C+\tan C\tan A$$
Multiplying by $\cos A \cos B \cos C=\frac 13$, we get $$\frac 13S=\sin A\sin B\cos C+\cos A\sin B\sin C+\sin A\cos B \sin C$$
However,$$\cos(A+B+C)=-1=\cos A\cos B\cos C-\sin A\sin B\cos C-\sin A\cos B\sin C-\cos A\sin B\sin C$$
Therefore, $$\frac 13S=\cos A\cos B\cos C+1=\frac 43\Rightarrow S=4$$
