1. In $\triangle A B C$, $\angle C>90^{\circ}, \tan A \cdot \tan B$ satisfies (  one ).
A. $\tan A \cdot \tan B>1$
B. $\tan A \cdot \tan B<1$
C. $\tan A \cdot \tan B=-1$
D. Cannot be determined
1. B. Since $A+B+C=180^{\circ}, C>90^{\circ}$, thus $A+B<90^{\circ}$, then $A<90^{\circ}-B<90^{\circ}$, therefore $\tan A<\tan \left(90^{\circ}-B\right)=\cot B$. So $\tan A \cdot \tan B<1$.