2. In $\triangle A B C$, the sides $a, b, c$ opposite to $\angle A, \angle B, \angle C$ form a geometric sequence. Then
$$
\frac{\sin A \cdot \cot C+\cos A}{\sin B \cdot \cot C+\cos B}
$$

the range of values is ( ).
(A) $(0,+\infty)$
(B) $\left(\frac{\sqrt{5}-1}{2}, \frac{\sqrt{5}+1}{2}\right)$
(C) $\left(0, \frac{\sqrt{5}+1}{2}\right)$
(D) $\left(\frac{\sqrt{5}-1}{2},+\infty\right)$
2. B.

Let the common ratio of $a, b, c$ be $q$.
Then $b=a q, c=a q^{2}$.
$$
\begin{array}{l}
\text { Hence } \frac{\sin A \cdot \cot C+\cos A}{\sin B \cdot \cot C+\cos B} \\
=\frac{\sin A \cdot \cos C+\cos A \cdot \sin C}{\sin B \cdot \cos C+\cos B \cdot \sin C} \\
=\frac{\sin (A+C)}{\sin (B+C)}=\frac{\sin B}{\sin A}=\frac{b}{a}=q . \\
\text { From }\left\{\begin{array} { l } 
{ a + b > c , } \\
{ b + c > a }
\end{array} \Rightarrow \left\{\begin{array}{l}
a+a q>a q^{2}, \\
a q+a q^{2}>a
\end{array}\right.\right. \\
\Rightarrow\left\{\begin{array}{l}
q^{2}-q-1<0
\end{array}\right. \\
\Rightarrow \frac{\sqrt{5}-1}{2}<q<\frac{\sqrt{5}+1}{2} .
\end{array}
$$