6 Let 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) $\left(\frac{\sqrt{5}-1}{2},+\infty\right)$
6 Let the common ratio of $a, b, c$ be $q$, then $b=a q, c=a q^{2}$, and
$$
\begin{aligned}
\frac{\sin A \cot C+\cos A}{\sin B \cot C+\cos B} & =\frac{\sin A \cos C+\cos A \sin C}{\sin B \cos C+\cos B \sin C} \\
& =\frac{\sin (A+C)}{\sin (B+C)} \\
& =\frac{\sin (\pi-B)}{\sin (\pi-A)} \\
& =\frac{\sin B}{\sin A} \\
& =\frac{b}{a} \\
& =q .
\end{aligned}
$$

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 $c$. For $a, b, c$ to form the sides of a triangle, it is necessary and sufficient that $a+b>c$ and $b+c>a$. Thus, we have the system of inequalities
$$
\left\{\begin{array}{l}
a+a q>a q^{2}, \\
a q+a q^{2}>a
\end{array}\right.
$$

which simplifies to
$$
\left\{\begin{array}{l}
q^{2}-q-1<0 \\
q^{2}+q-1>0
\end{array}\right.
$$

Solving these, we get
$$
\left\{\begin{array}{l}
\frac{1-\sqrt{5}}{2}<q<\frac{1+\sqrt{5}}{2} \\
q>\frac{\sqrt{5}-1}{2} \text { or } q<-\frac{\sqrt{5}+1}{2}
\end{array}\right.
$$

Thus,
$$
\frac{\sqrt{5}-1}{2}<q<\frac{\sqrt{5}+1}{2}
$$

Therefore, the required range is $\left(\frac{\sqrt{5}-1}{2}, \frac{\sqrt{5}+1}{2}\right)$. Hence, the answer is C.