8. Given $\sin (\alpha+\beta)=\frac{1}{2}, \sin (\alpha-\beta)=\frac{1}{3}, \tan \alpha \cdot \cot \beta=$
8. 5. $\tan \alpha \cdot \cot \beta=\frac{\sin \alpha \cos \beta}{\cos \alpha \sin \beta}=\frac{\frac{1}{2}[\sin (\alpha+\beta)+\sin (\alpha-\beta)]}{\frac{1}{2}[\sin (\alpha+\beta)-\sin (\alpha-\beta)]}=\frac{\frac{1}{2}+\frac{1}{3}}{\frac{1}{2}-\frac{1}{3}}=5$.