2. If the sides $a, b, c$ opposite to the internal angles $A, B, C$ of $\triangle A B C$ form a geometric sequence, then the range of $\frac{\sin A \cot C+\cos A}{\sin B \cot C+\cos B}$ is
A. $(0,+\infty)$
B. $\left(0, \frac{\sqrt{5}+1}{2}\right)$
C. $\left(\frac{\sqrt{5}-1}{2}, \frac{\sqrt{5}+1}{2}\right)$
D. $\left(\frac{\sqrt{5}-1}{2},+\infty\right)$
Solve:
$$
\frac{\sin A \cot C+\cos A}{\sin B \cot C+\cos B}=\frac{\sin A \cos C+\cos A \sin C}{\sin B \cos C+\cos B \sin C}=\frac{\sin B}{\sin A}=\frac{b}{a}=q \text {, }
$$

and $\left\{\begin{array}{l}a(1+q)>a q^{2}, \\ a\left(q+q^{2}\right)>a\end{array} \Rightarrow q \in\left(\frac{\sqrt{5}-1}{2}, \frac{\sqrt{5}+1}{2}\right)\right.$, so the answer is $C$.