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

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

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$\frac{56}{33}$

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(B) Given $\cos(\alpha + \beta) = \frac{4}{5}$. Since $0 \le \alpha, \beta \le \frac{\pi}{4}$,we have $0 \le \alpha + \beta \le \frac{\pi}{2}$,so $	an(\alpha + \beta) = \frac{3}{4}$.Given $\sin(\alpha - \beta) = \frac{5}{13}$. Since $0 \le \alpha, \beta \le \frac{\pi}{4}$,we have $-\frac{\pi}{4} \le \alpha - \beta \le \frac{\pi}{4}$,so $	an(\alpha - \beta) = \frac{5}{12}$.We know that $2\alpha = (\alpha + \beta) + (\alpha - \beta)$.Using the formula $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,we get:$	an 2\alpha = 	an((\alpha + \beta) + (\alpha - \beta)) = \frac{	an(\alpha + \beta) + 	an(\alpha - \beta)}{1 - 	an(\alpha + \beta)	an(\alpha - \beta)}$$	an 2\alpha = \frac{\frac{3}{4} + \frac{5}{12}}{1 - (\frac{3}{4} 	imes \frac{5}{12})} = \frac{\frac{9+5}{12}}{1 - \frac{15}{48}} = \frac{\frac{14}{12}}{\frac{33}{48}} = \frac{14}{12} 	imes \frac{48}{33} = \frac{14 	imes 4}{33} = \frac{56}{33}$.

If $\cos(\alpha + \beta) = \frac{4}{5}$ and $\sin(\alpha - \beta) = \frac{5}{13}$,where $0 \le \alpha, \beta \le \frac{\pi}{4}$. Then $\tan 2\alpha = $

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