3. In $\triangle A B C$, $B C=3$, the inradius $r$ $=\frac{\sqrt{3}}{2}$. Then the value of $\operatorname{ctg} \frac{B}{2}+\operatorname{ctg} \frac{C}{2}$ is ( ).
(A) $\frac{\sqrt{3}}{2}$
(B) $\frac{2}{\sqrt{3}}$
(C) $\frac{3 \sqrt{3}}{2}$
(D) $2 \sqrt{3}$
3. (D).

As shown in the right figure, the incenter $I$ of the triangle is the intersection of the three angle bisectors, so $\operatorname{ctg} \frac{B}{2}+\operatorname{ctg} \frac{C}{2}$ $=\frac{B D}{I D}+\frac{C D}{I D}=\frac{B C}{r}=2 \sqrt{3}$.