3. Given $\triangle A B C, \tan A, \tan B, \tan C$ are all integers, and $\angle A>\angle B>\angle C$. Then $\tan B=$ $\qquad$
3. 2 .

Since $\angle A>\angle B>\angle C$, therefore, $\angle B, \angle C$ are acute angles.
Also, $\tan B, \tan C$ are positive integers, then
$$
\tan (\pi-A)=-\frac{\tan B+\tan C}{\tan B \cdot \tan C-1}<0 .
$$

Thus, $\angle A$ is also an acute angle.
At this point, $\tan C \geqslant 1, \tan B \geqslant 2, \tan A \geqslant 3$.
Therefore, $\frac{\tan A+\tan B}{\tan A \cdot \tan B-1}=\tan C \geqslant 1$.
Consequently, $(\tan A - (\tan B-1)) \leqslant 2$.
And $\tan A-1 \geqslant 2, \tan B-1 \geqslant 1$, from this we know $\tan A=3, \tan B=2, \tan C=1$.