$3 \triangle ABC$ is inscribed in a unit circle, and the angle bisectors of the three interior angles $A, B, C$ are extended to intersect this circle at $A_{1}, B_{1}, C_{1}$. Then
$$
\frac{A A_{1} \cdot \cos \frac{A}{2} + B B_{1} \cdot \cos \frac{B}{2} + C C_{1} \cdot \cos \frac{C}{2}}{\sin A + \sin B + \sin C}
$$

is ( ).
(A) 2
(B) 4
(C) 6
(D) 8
Solve as shown in the figure, connect $B A_{1}$, then
$$
\begin{aligned}
A A_{1} & =2 \sin \left(B+\frac{A}{2}\right) \\
& =2 \sin \left(\frac{A+B+C}{2}+\frac{B}{2}-\frac{C}{2}\right) \\
& =2 \cos \left(\frac{B}{2}-\frac{C}{2}\right) .
\end{aligned}
$$

Therefore,
$$
\begin{aligned}
A A_{1} \cdot \cos \frac{A}{2} & =2 \cos \left(\frac{B}{2}-\frac{C}{2}\right) \cos \frac{A}{2} \\
& =\cos \frac{A+B-C}{2}+\cos \frac{A+C-B}{2} \\
& =\cos \left(\frac{\pi}{2}-C\right)+\cos \left(\frac{\pi}{2}-B\right) \\
& =\sin C+\sin B,
\end{aligned}
$$

Similarly,
Therefore,
$$
\begin{array}{c}
A A_{1} \cdot \cos \frac{A}{2}+B B_{1} \cdot \cos \frac{B}{2}+C C_{1} \cdot \cos \frac{C}{2} \\
=2(\sin A+\sin B+\sin C),
\end{array}
$$

Thus, the original expression $=\frac{2(\sin A+\sin B+\sin C)}{\sin A+\sin B+\sin C}=2$, hence the answer is A.