If (alpha+beta) is not a multiple of frac{pi} | vedclass

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(A) Given $3 \sin (\alpha-\beta)=5 \cos (\alpha+\beta)$.Expanding the terms,we get $3(\sin \alpha \cos \beta - \cos \alpha \sin \beta) = 5(\cos \alpha

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(A) Given $3 \sin (\alpha-\beta)=5 \cos (\alpha+\beta)$.Expanding the terms,we get $3(\sin \alpha \cos \beta - \cos \alpha \sin \beta) = 5(\cos \alpha \cos \beta - \sin \alpha \sin \beta)$.Rearranging the terms,we have $\sin \alpha(3 \cos \beta + 5 \sin \beta) = \cos \alpha(5 \cos \beta + 3 \sin \beta)$.Thus,$	an \alpha = \frac{5 \cos \beta + 3 \sin \beta}{3 \cos \beta + 5 \sin \beta} = \frac{5 + 3 	an \beta}{3 + 5 	an \beta}$.Now,$	an \left(\frac{\pi}{4} + \alphaight) = \frac{1 + 	an \alpha}{1 - 	an \alpha} = \frac{1 + \frac{5 + 3 	an \beta}{3 + 5 	an \beta}}{1 - \frac{5 + 3 	an \beta}{3 + 5 	an \beta}} = \frac{3 + 5 	an \beta + 5 + 3 	an \beta}{3 + 5 	an \beta - 5 - 3 	an \beta} = \frac{8 + 8 	an \beta}{-2 + 2 	an \beta} = \frac{8(1 + 	an \beta)}{-2(1 - 	an \beta)} = -4 	an \left(\frac{\pi}{4} + \betaight)$.Therefore,$	an \left(\frac{\pi}{4} + \alphaight) + 4 	an \left(\frac{\pi}{4} + \betaight) = 0$.

If $(\alpha+\beta)$ is not a multiple of $\frac{\pi}{2}$ and $3 \sin (\alpha-\beta)=5 \cos (\alpha+\beta)$,then $\tan \left(\frac{\pi}{4}+\alpha\right)+4 \tan \left(\frac{\pi}{4}+\beta\right)=$

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