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

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(B) Given $\cos (\alpha + \beta ) = \frac{4}{5}$ and $\sin (\alpha - \beta ) = \frac{5}{13}$.Since $0 < \alpha, \beta < \frac{\pi}{4}$,we have $0 < \a

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$\frac{56}{33}$

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(B) Given $\cos (\alpha + \beta ) = \frac{4}{5}$ and $\sin (\alpha - \beta ) = \frac{5}{13}$.Since $0 Thus,$\sin (\alpha + \beta ) = \sqrt{1 - (\frac{4}{5})^2} = \frac{3}{5}$ and $\cos (\alpha - \beta ) = \sqrt{1 - (\frac{5}{13})^2} = \frac{12}{13}$.We know that $2\alpha = (\alpha + \beta ) + (\alpha - \beta )$.Using the formula $\sin(A+B) = \sin A \cos B + \cos A \sin B$:$\sin 2\alpha = \sin ((\alpha + \beta ) + (\alpha - \beta )) = \sin (\alpha + \beta ) \cos (\alpha - \beta ) + \cos (\alpha + \beta ) \sin (\alpha - \beta )$.$\sin 2\alpha = (\frac{3}{5} 	imes \frac{12}{13}) + (\frac{4}{5} 	imes \frac{5}{13}) = \frac{36}{65} + \frac{20}{65} = \frac{56}{65}$.Using the formula $\cos(A+B) = \cos A \cos B - \sin A \sin B$:$\cos 2\alpha = \cos ((\alpha + \beta ) + (\alpha - \beta )) = \cos (\alpha + \beta ) \cos (\alpha - \beta ) - \sin (\alpha + \beta ) \sin (\alpha - \beta )$.$\cos 2\alpha = (\frac{4}{5} 	imes \frac{12}{13}) - (\frac{3}{5} 	imes \frac{5}{13}) = \frac{48}{65} - \frac{15}{65} = \frac{33}{65}$.Therefore,$	an 2\alpha = \frac{\sin 2\alpha}{\cos 2\alpha} = \frac{56/65}{33/65} = \frac{56}{33}$.

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

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