3. Given an acute triangle $\triangle A B C$. The following judgments are given:
(1) Three segments of lengths $\sin 2 A 、 \sin 2 B 、 \sin 2 C$ can definitely form a triangle;
(2) Three segments of lengths $\cos A 、 \cos B 、 \cos C$ can definitely form a triangle;
(3) Three segments of lengths $\cos A 、 \cos B 、 \sin C$ can definitely form a triangle;
(4) Three segments of lengths $\tan A 、 \tan B 、 \tan C$ can definitely form a triangle.
Among them, the correct judgments are ( ) in number.
(A) 4
(B) 3
(C) 2
(D) 1
3.C.
$$
\begin{array}{l}
\sin 2 A+\sin 2 B-\sin 2 C \\
=2 \sin (A+B) \cdot \cos (A-B)-2 \sin C \cdot \cos C \\
=2 \sin C[\cos (A-B)+\cos (A+B)] \\
=4 \cos A \cdot \cos B \cdot \sin C>0 .
\end{array}
$$

Similarly, $\sin 2 A+\sin 2 C-\sin 2 B>0$,
$$
\sin 2 B+\sin 2 C-\sin 2 A>0 \text {. }
$$

Therefore, (1) is correct.
For (2), consider the extreme case: $\angle A \rightarrow 90^{\circ}, \angle B \rightarrow 90^{\circ}$, $\angle C \rightarrow 0^{\circ}$, at this time, $\cos A \rightarrow 0, \cos B \rightarrow 0, \cos C \rightarrow 1$ does not satisfy $\cos A+\cos B>\cos C$. Therefore, (2) is incorrect.
$$
\begin{array}{l}
\cos A+\cos B-\sin C \\
=2 \cos \frac{A+B}{2} \cdot \cos \frac{A-B}{2}-2 \sin \frac{C}{2} \cdot \cos \frac{C}{2} \\
=2 \sin \frac{C}{2}\left(\cos \frac{A-B}{2}-\cos \frac{C}{2}\right) \\
=4 \sin \frac{C}{2} \cdot \sin \frac{\pi-2 B}{4} \cdot \sin \frac{\pi-2 A}{4}>0 .
\end{array}
$$

Similarly, $\cos A+\sin C-\cos B>0$, $\cos B+\sin C-\cos A>0$.
Therefore, (3) is correct.
For (4), consider a counterexample: $\angle A=45^{\circ}, \angle B=60^{\circ}, \angle C$ $=75^{\circ}$, at this time, $\tan A=1, \tan B=\sqrt{3}, \tan C=2+\sqrt{3}$, $\tan A+\tan B<\tan C$.