1. Given $\alpha+\beta=15^{\circ}$. Then $\frac{1-\operatorname{tg} \alpha-\operatorname{tg} \beta-\operatorname{tg} \alpha \operatorname{tg} \beta}{1+\operatorname{tg} \alpha+\operatorname{tg} \beta-\operatorname{tg} \alpha \operatorname{tg} \beta}$
\[
\begin{array}{l}
=1 \cdot \frac{\sqrt{3}}{3} \text {. } \\ 
\text { Original expression }=\frac{1-\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}}{1+\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}}=\frac{1-\tan(\alpha+\beta)}{1+\tan(\alpha+\beta)} \\ 
=\frac{1-\tan 15^{\circ}}{1+\tan 15^{\circ}}=\tan 30^{\circ}=\frac{\sqrt{3}}{3} .
\end{array}
\]