2. Given $\tan (3 \alpha-2 \beta)=\frac{1}{2}, \tan (5 \alpha-4 \beta)=\frac{1}{4}$, then $\tan \alpha=$
Solution: $\frac{13}{16}$
By the double angle formula
$$
\tan (6 \alpha-4 \beta)=\frac{2 \tan (3 \alpha-2 \beta)}{1-\tan ^{2}(3 \alpha-2 \beta)}=\frac{4}{3}
$$

By the sum and difference formula
(b;
$$
\tan \alpha=\tan [(6 \alpha-4 \beta)-(5 \alpha-4 \beta)]=\frac{\tan (6 \alpha-4 \beta)-\tan (5 \alpha-4 \beta)}{1+\tan (6 \alpha-4 \beta) \tan (5 \alpha-4 \beta)}=\frac{13}{16}
$$