1. If $S_{\triangle A B C}=\frac{1}{4} b^{2} \tan B$, where $b$ is the side opposite to $\angle B$, then $\frac{\cot B}{\cot A+\cot C}=$ $\qquad$ .
$\begin{array}{l}\text { II. 1. } \frac{1}{2} . \\ S_{\triangle A B C}=\frac{1}{4} b^{2} \tan B \Leftrightarrow \frac{1}{2} a c \sin B=\frac{1}{4} b^{2} \tan B \\ \Leftrightarrow 2 \cot B=\frac{\sin (A+C)}{\sin A \cdot \sin C}=\cot A+\cot C \\ \Leftrightarrow \frac{\cot B}{\cot A+\cot C}=\frac{1}{2} .\end{array}$