5. Let $\alpha, \beta$ satisfy $\tan \left(\alpha+\frac{\pi}{3}\right)=-3, \tan \left(\beta-\frac{\pi}{6}\right)=5$, then the value of $\tan (\alpha-\beta)$ is
Answer: $-\frac{7}{4}$.
Solution: According to the tangent formula for the difference of two angles, we have $\tan \left(\left(\alpha+\frac{\pi}{3}\right)-\left(\beta-\frac{\pi}{6}\right)\right)=\frac{-3-5}{1+(-3) \times 5}=\frac{4}{7}$, which means $\tan \left(\alpha-\beta+\frac{\pi}{2}\right)=\frac{4}{7}$. Therefore, $\tan (\alpha-\beta)=-\cot \left(\alpha-\beta+\frac{\pi}{2}\right)=-\frac{7}{4}$.