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. 

Rephrasing Mathematics's answer:
Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$ and $\cos(A+B)=\cdots=-\cos C$
$\cos^2A+\cos^2B+\cos^2C=1+\cos^2A-\sin^2B+\cos^2C=1-2\cos A\cos B\cos C$
If $\cos A\cos B\cos C=S$
$\iff\cos^2A+\cos^2B+\cos^2C=1-2S$
let $y=\tan A\tan B\iff y-1=-\dfrac{\cos(A+B)}{\cos A\cos B}=\dfrac{\cos^2C}S$
$\iff\cos^2C=S(y-1)$
$\sum S(y-1)=\sum\cos^2C=1-2S\iff y=3+\dfrac{1-2S}S$
Here $S=\dfrac13$
