Let cos (alpha+beta)=frac{4}{5} and sin (alph | vedclass

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(B) Given $\cos (\alpha+\beta)=\frac{4}{5}$. Since $0 \leq \alpha, \beta \leq \frac{\pi}{4}$,we have $0 \leq \alpha+\beta \leq \frac{\pi}{2}$,so $	an

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$\frac{56}{33}$

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(B) Given $\cos (\alpha+\beta)=\frac{4}{5}$. Since $0 \leq \alpha, \beta \leq \frac{\pi}{4}$,we have $0 \leq \alpha+\beta \leq \frac{\pi}{2}$,so $	an (\alpha+\beta)=\frac{3}{4}$.Given $\sin (\alpha-\beta)=\frac{5}{13}$. Since $0 \leq \alpha, \beta \leq \frac{\pi}{4}$,we have $-\frac{\pi}{4} \leq \alpha-\beta \leq \frac{\pi}{4}$,so $	an (\alpha-\beta)=\frac{5}{12}$.We know that $2\alpha = (\alpha+\beta)+(\alpha-\beta)$.Therefore,$	an 2 \alpha = 	an \{(\alpha+\beta)+(\alpha-\beta)\} = \frac{	an (\alpha+\beta)+	an (\alpha-\beta)}{1-	an (\alpha+\beta) \cdot 	an (\alpha-\beta)}$.Substituting the values: $	an 2 \alpha = \frac{\frac{3}{4}+\frac{5}{12}}{1-(\frac{3}{4} \cdot \frac{5}{12})} = \frac{\frac{9+5}{12}}{1-\frac{15}{48}} = \frac{\frac{14}{12}}{\frac{33}{48}} = \frac{14}{12} \cdot \frac{48}{33} = \frac{14 \cdot 4}{33} = \frac{56}{33}$.

Let $\cos (\alpha+\beta)=\frac{4}{5}$ and $\sin (\alpha-\beta)=\frac{5}{13}$,where $0 \leq \alpha, \beta \leq \frac{\pi}{4}$,then $\tan 2 \alpha=$

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