(2) In a scalene $\triangle A B C$, angle $B=60^{\circ}$, then the positional relationship between the line $l_{1}: x+\frac{\sin A+\sin C}{\sqrt{3}} y+1=0$ and the line $l_{2}: x \cdot \tan \left(60^{\circ}-C\right)+y \cdot(\sin A-\sin C)-\tan \frac{C-A}{2}=0$ is ( ).
(A) Perpendicular
(B) Parallel
(C) Coincident
(D) Other cases
2 Since $A+C=120^{\circ}$, then
$$
\begin{array}{l}
\tan \left(60^{\circ}-C\right)=\tan \left(\frac{A+C}{2}-C\right)=\tan \frac{A-C}{2}, \\
\sin A-\sin C=2 \sin \frac{A-C}{2} \cos \frac{A+C}{2}=\sin \frac{A-C}{2}, \\
\sin A+\sin C=2 \sin \frac{A+C}{2} \cos \frac{A-C}{2}=\sqrt{3} \cos \frac{A-C}{2} .
\end{array}
$$

Therefore, the equations of lines $l_{1}$ and $l_{2}$ can both be simplified to
$$
x+y \cos \frac{A-C}{2}+1=0,
$$

Thus, the two lines coincide. Hence, the answer is C.