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. 

Hint: For a triangle ABC
$A+B=\pi-C$
and 
$1-2\cos A \cos B \cos C=\cos^2A+\cos^2B+\cos^2C$
Edit $1$:
$A+B=\pi-C$
Apply $\cos$ on both sides
Divide each term by $\cos A.\cos B$
We get $\tan A.\tan B=1+\frac{\cos C}{\cos A. \cos B}$
Similarly write $2$ more equations and add three equations.
Now use $1-2\cos A \cos B \cos C=\cos^2A+\cos^2B+\cos^2C$
