$\begin{array}{l}\text { 1. In } \triangle A B C \text {, if } \\ \tan A \cdot \tan B=\tan A \cdot \tan C+\tan C \cdot \tan B \text {, } \\ \text { then } \frac{a^{2}+b^{2}}{c^{2}}=\end{array}$
$-1.3$.
Given the equation can be transformed as
$$
\begin{array}{l}
\frac{\sin A \cdot \sin B}{\cos A \cdot \cos B}=\frac{\sin A \cdot \sin C}{\cos A \cdot \cos C}+\frac{\sin B \cdot \sin C}{\cos B \cdot \cos C} \\
\Rightarrow \sin A \cdot \sin B \cdot \cos C \\
\quad=\sin C \cdot(\sin A \cdot \cos B+\cos A \cdot \sin B) \\
\quad=\sin ^{2} C \\
\Rightarrow a b \frac{a^{2}+b^{2}-c^{2}}{2 a b}=c^{2} \\
\Rightarrow a^{2}+b^{2}=3 c^{2} .
\end{array}
$$

Therefore, the answer is 3.