3. In the acute triangle $\triangle A B C$, $\angle A: \angle B, \angle C$ are its interior angles. Let $T=\operatorname{ctg} 2 A+\operatorname{ctg} 2 B+\operatorname{ctg} 2 C$. Then it must be ( ).
(A.) T $>0$
(B) $\cdots \geqslant 0$
(c) $T<0$
(D) $T \leqslant 0$
3. (C).

From $\angle A + \angle B > \frac{\pi}{2}$, we get $\angle A > \frac{\pi}{2} - \angle B$, which means $\operatorname{tg} A > \operatorname{tg}\left(\frac{\pi}{2} - \angle B\right) = \operatorname{ctg} B$.
Similarly, $\operatorname{tg} B > \operatorname{ctg} C, \operatorname{tg} C > \operatorname{ctg} A$. Therefore,
$$
\begin{array}{l}
\operatorname{tg} A + \operatorname{tg} B + \operatorname{tg} C > \operatorname{ctg} A + \operatorname{ctg} B + \operatorname{ctg} C \\
\operatorname{ctg} 2 A + \operatorname{ctg} 2 B + \operatorname{ctg} 2 C \\
= \frac{\cos^2 A - \sin^2 A}{2 \sin A \cos A} + \operatorname{ctg} 2 B + \operatorname{ctg} 2 C \\
= \frac{1}{2} (\operatorname{ctg} A - \operatorname{tg} A + \operatorname{ctg} B - \operatorname{tg} B + \operatorname{ctg} C - \operatorname{tg} C) \\
< 0
\end{array}
$$