2. If in $\triangle A B C$, $\tan A, \tan B, \tan C$ are all integers, and $A>B>C$, then $\tan B=$
2.2

Analysis: Since $\tan A+\tan B+\tan C=\tan A \tan B \tan C$, and $A>B>C$, then when $A\tan B>\tan C>0$, thus $3 \tan A>\tan A+\tan B+\tan C=\tan A \tan B \tan C$, which means $\tan B \tan C \leq 2$. If $\tan B=2, \tan C=1$, then $\tan A=3$, satisfying the condition; if $\tan B=\tan C=1$, then $\tan A$ has no solution; when $A>\frac{\pi}{2}$, $\tan A\tan C>0$, then $\tan A=\frac{\tan B+\tan C}{\tan B \tan C-1}<0$, meaning $\tan B \tan C<1$, no solution.