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

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Question

(A) Given that $\cos (\alpha+\beta) = \frac{4}{5}$. Since $0 < \alpha, \beta < \frac{\pi}{4}$,we have $0 < \alpha+\beta < \frac{\pi}{2}$. Thus,$	an (

Vedclass · Accepted Answer

$\frac{56}{33}$

Vedclass · Answer

(A) Given that $\cos (\alpha+\beta) = \frac{4}{5}$. Since $0 Given $\sin (\alpha-\beta) = \frac{5}{13}$. Since $0 Now,$	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{3}{4} + \frac{5}{12}}{1 - (\frac{3}{4} \cdot \frac{5}{12})} = \frac{\frac{9+5}{12}}{1 - \frac{15}{48}} = \frac{14/12}{33/48} = \frac{14}{12} \cdot \frac{48}{33} = \frac{14 \cdot 4}{33} = \frac{56}{33}$.

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

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