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 find $	an(\alpha + \beta) = \frac{\sqrt{1 - (3/5)^2}}{3/5} = \frac{4/5}{3/5} = \frac{4}{3}$.Since $\sin(\alpha - \beta) = \frac{5}{13}$,we find $	an(\alpha - \beta) = \frac{5/13}{\sqrt{1 - (5/13)^2}} = \frac{5/13}{12/13} = \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} \cdot \frac{5}{12})} = \frac{\frac{16+5}{12}}{1 - \frac{20}{36}} = \frac{21/12}{16/36} = \frac{21}{12} \cdot \frac{36}{16} = \frac{21 \cdot 3}{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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