14. Given $\tan \frac{\alpha+\beta}{2}=\frac{\sqrt{6}}{2}, \cot \alpha \cdot \cot \beta=\frac{7}{13}$, find the value of $\cos (\alpha-\beta)$.
14. Since $\cos (\alpha+\beta)=\frac{1-\tan ^{2} \frac{\alpha+\beta}{2}}{1+\tan ^{2} \frac{\alpha+\beta}{2}}=-\frac{1}{5}$, then $\cos \alpha \cos \beta-\sin \alpha \sin \beta=-\frac{1}{5}$, $\cot \alpha \cos \beta=\frac{\cos \alpha \cdot \cos \beta}{\sin \alpha \sin \beta}=\frac{7}{13}$, solving gives $\sin \alpha \sin \beta=\frac{13}{30}, \cos \alpha \cos \beta=\frac{7}{30}$. Therefore, $\cos (\alpha-\beta)=$ $\cos \alpha \cos \beta+\sin \alpha \sin \beta=\frac{2}{3}$.