6. 若 $\cos (\alpha+\beta)=\frac{1}{5}, \cos (\alpha-\beta)=\frac{3}{5}$, 则 $\tan \alpha \cdot \tan \beta=$
知, 得 $\left\{\begin{array}{l}\cos \alpha+\beta=\frac{1}{5}, \\ \cos \alpha-\beta=\frac{3}{5}\end{array}\right.$

即 $\left\{\begin{array}{l}\cos \alpha \cos \beta-\sin \alpha \sin \beta=\frac{1}{5}, \\ \cos \alpha \cos \beta+\sin \alpha \sin \beta=\frac{3}{5} .\end{array}\right.$

解得 $\left\{\begin{array}{l}\cos \alpha \cdot \cos \beta=\frac{2}{5}, \\ \sin \alpha \cdot \sin \beta=\frac{1}{5} .\end{array}\right.$

所以 $\tan \alpha \cdot \tan \beta=\frac{\sin \alpha \sin \beta}{\cos \alpha \cos \beta}=\frac{1}{2}$.