If cos (alpha + beta) = frac{3}{5},sin (alpha | vedclass

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(C) Given $0 < \alpha, \beta < \frac{\pi}{4}$,we have $0 < \alpha + \beta < \frac{\pi}{2}$ and $-\frac{\pi}{4} < \alpha - \beta < \frac{\pi}{4}$.Since

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$\frac{63}{16}$

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(C) Given $0 Since $\cos(\alpha + \beta) = \frac{3}{5}$,we have $	an(\alpha + \beta) = \frac{4}{3}$.Since $\sin(\alpha - \beta) = \frac{5}{13}$,we have $\cos(\alpha - \beta) = \sqrt{1 - (\frac{5}{13})^2} = \frac{12}{13}$,so $	an(\alpha - \beta) = \frac{5}{12}$.Now,$	an(2\alpha) = 	an((\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) = \frac{\frac{4}{3} + \frac{5}{12}}{1 - (\frac{4}{3} 	imes \frac{5}{12})} = \frac{\frac{16+5}{12}}{1 - \frac{20}{36}} = \frac{\frac{21}{12}}{\frac{16}{36}} = \frac{7}{4} 	imes \frac{36}{16} = \frac{7 	imes 9}{16} = \frac{63}{16}$.

If $\cos (\alpha + \beta) = \frac{3}{5}$,$\sin (\alpha - \beta) = \frac{5}{13}$ and $0 < \alpha, \beta < \frac{\pi}{4}$,then $\tan (2\alpha)$ is equal to

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