17. Given $\cos ^{2}(\alpha-\beta)-\cos ^{2}(\alpha+\beta)=\frac{1}{2},(1+\cos 2 \alpha)(1+\cos 2 \beta)=\frac{1}{3}$, find the value of $\tan \alpha \tan \beta$.
17. $\cos ^{2}(\alpha-\beta)-\cos ^{2}(\alpha+\beta)=[\cos \alpha \cos \beta+\sin \alpha \sin \beta]^{2}-[\cos \alpha \cos \beta-\sin \alpha \sin \beta]^{2}=$ $4 \sin \alpha \sin \beta \cos \alpha \cos \beta,(1+\cos 2 \alpha)(1+\cos 2 \beta)=4 \cos ^{2} \alpha \cos ^{2} \beta$. Dividing the two equations yields $\frac{\cos ^{2}(\alpha-\beta)-\cos ^{2}(\alpha+\beta)}{(1+\cos 2 \alpha)(1+\cos 2 \beta)}=\tan \alpha \cdot \tan \beta$, so $\tan \alpha \cdot \tan \beta=\frac{3}{2}$.