Given that \(\cos (\alpha+\beta)=\frac{4}{5}\) and \(\sin (\alpha-\beta)=\frac{5}{13}\) Thus \(\tan (\alpha+\beta)=\frac{1}{4}\) and \(\tan (\alpha-\beta)=\frac{5}{12}\) \[\] Now \(\tan 2 \alpha=\tan ((\alpha+\beta)+(\alpha-\beta))\) \[\] \(=\frac{\tan (\alpha+\beta)+\tan (\alpha-\beta)}{1-\tan (\alpha+\beta) \tan (\alpha-\beta)}\) \[\] \(=\frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4} \times \frac{5}{12}}=\frac{\frac{27+15}{36}}{1-\frac{5}{16}}=\frac{\frac{42}{36}}{\frac{11}{16}}=\frac{56}{33}\) \[\] Choice (A)\[\]