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Let $\cos (\alpha+\beta)=\frac{4}{5}$ and let $\sin (\alpha-\beta)=\frac{5}{13}$, where $0 \leq \alpha, \beta \leq \frac{\pi}{4}$, then $\tan 2 \alpha=$
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Verified Answer
The correct answer is:
$\frac{56}{33}$
$\frac{56}{33}$
$\begin{array}{ll}\cos (\alpha+\beta)=\frac{4}{5} & \Rightarrow \tan (\alpha+\beta)=\frac{3}{4} \\ \sin (\alpha-\beta)=\frac{5}{13} & \Rightarrow \tan (\alpha-\beta)=\frac{5}{12} \\ \tan 2 \alpha=\tan (\alpha+\beta+\alpha-\beta)=\frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4} \frac{5}{12}}=\frac{56}{33}\end{array}$
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