If \( \cos (\alpha+\beta)=\frac{4}{5} \) and \( \sin (\alpha-\beta)=\frac{5}{13}, 0<\alpha, \beta<\frac{\pi}{4} \) , then \( \tan (2 \alpha)= \)
\(\alpha + \beta\) is acute with \(\cos = \frac{4}{5}\), so \(\tan(\alpha + \beta) = \frac{3}{4}\).
\(\sin(\alpha - \beta) = \frac{5}{13}\), so \(\tan(\alpha - \beta) = \frac{5}{12}\).
\(\tan 2\alpha = \tan[(\alpha + \beta) + (\alpha - \beta)] = \frac{\frac{3}{4} + \frac{5}{12}}{1 - \frac{3}{4} \cdot \frac{5}{12}} = \frac{14/12}{33/48} = \frac{56}{33}\)
Answer: (A) \(\frac{56}{33}\)
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