$$\frac{\sin A\cot C+\cos A}{\sin B\cot C+\cos B}=\frac{\sin A\cos C+\sin C\cos A}{\sin B\cos C+\sin C\cos B}=\frac{\sin \left( A+C \right)}{\sin \left( B+C \right)}=\frac{\sin B}{\sin A}=\frac{b}{a}$$，  由题设有$${{b}^{2}}=ac$$，则$$a\left\textbar{} a-b \right\textbar\textless{}{{b}^{2}}\textless{}a\left( a+b \right)$$，解得$$\frac{\sqrt{5}-1}{2}\textless{}\frac{b}{a}\textless{}\frac{\sqrt{5}+1}{2}$$．  故选$$\text{C}$$． 
若$$\triangle ABC$$的内角$$A$$，$$B$$，$$C$$所对的边$$a$$，$$b$$，$$c$$成等比数列，则$$\frac{\sin A\cot C+\cos A}{\sin B\cot C+\cos B}$$的取值范围是（~ ~ ）．

1: $$\left( 0,+\infty \right)$$ 
2: $$\left( 0,\frac{\sqrt{5}+1}{2} \right)$$ 
3: $$\left( \frac{\sqrt{5}-1}{2},\frac{\sqrt{5}+1}{2} \right)$$ 
4: $$\left( \frac{\sqrt{5}-1}{2},+\infty \right)$$ 