6. Let $\triangle A B C$ have $\angle A, \angle B, \angle C$ opposite sides $a, b, c$ in a geometric progression. Then the range of $\frac{\sin A \cdot \cot C+\cos A}{\sin B \cdot \cot C+\cos B}$ 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) $\left(\frac{\sqrt{5}-1}{2},+\infty\right)$
6. C.

Let the common ratio of $a, b, c$ be $q$. Then $b=a q, c=a q^{2}$.
$$
\begin{array}{l} 
\text { and } \frac{\sin A \cdot \cot C+\cos A}{\sin B \cdot \cot C+\cos B} \\
= \frac{\sin A \cdot \cos C+\cos A \cdot \sin C}{\sin B \cdot \cos C+\cos B \cdot \sin C} \\
=\frac{\sin (A+C)}{\sin (B+C)}=\frac{\sin (\pi-B)}{\sin (\pi-A)} \\
=\frac{\sin B}{\sin A}=\frac{b}{a}=q .
\end{array}
$$

Therefore, we only need to find the range of $q$.
Since $a, b, c$ form a geometric sequence, the largest side can only be $a$ or $c$. Therefore, for $a, b, c$ to form the sides of a triangle, it must and only must satisfy $a+b>c$ and $b+c>a$, leading to the system of inequalities
$$
\left\{\begin{array} { l } 
{ a + a q > a q ^ { 2 } , } \\
{ a q + a q ^ { 2 } > a , }
\end{array} \text { i.e., } \left\{\begin{array}{l}
q^{2}-q-1<0 .
\end{array}\right.\right.
$$

Solving this, we get $\left\{\begin{array}{l}\frac{1-\sqrt{5}}{2}<q<\frac{\sqrt{5}-1}{2} \text { or } q<-\frac{\sqrt{5}+1}{2} .\end{array}\right.$
Thus, $\frac{\sqrt{5}-1}{2}<q<\frac{\sqrt{5}+1}{2}$.
Therefore, the required range is $\left(\frac{\sqrt{5}-1}{2}, \frac{\sqrt{5}+1}{2}\right)$.