If sin (alpha+beta)=5 sin (alpha-beta),then f | vedclass

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(D) Given $\sin (\alpha+\beta)=5 \sin (\alpha-\beta)$.Using the expansion formula,we have $\sin \alpha \cos \beta + \cos \alpha \sin \beta = 5(\sin \a

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$	an (\alpha-\beta)$

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(D) Given $\sin (\alpha+\beta)=5 \sin (\alpha-\beta)$.Using the expansion formula,we have $\sin \alpha \cos \beta + \cos \alpha \sin \beta = 5(\sin \alpha \cos \beta - \cos \alpha \sin \beta)$.Rearranging the terms: $\sin \alpha \cos \beta + \cos \alpha \sin \beta = 5 \sin \alpha \cos \beta - 5 \cos \alpha \sin \beta$.$6 \cos \alpha \sin \beta = 4 \sin \alpha \cos \beta$.Dividing by $2 \cos \alpha \cos \beta$,we get $3 	an \beta = 2 	an \alpha$.Now,consider the expression $\frac{\sin 2 \beta}{5-\cos 2 \beta}$.Using the half-angle formulas $\sin 2 \beta = \frac{2 	an \beta}{1+	an^2 \beta}$ and $\cos 2 \beta = \frac{1-	an^2 \beta}{1+	an^2 \beta}$,we get:$\frac{\frac{2 	an \beta}{1+	an^2 \beta}}{5 - \frac{1-	an^2 \beta}{1+	an^2 \beta}} = \frac{2 	an \beta}{5(1+	an^2 \beta) - (1-	an^2 \beta)} = \frac{2 	an \beta}{5 + 5 	an^2 \beta - 1 + 	an^2 \beta} = \frac{2 	an \beta}{4 + 6 	an^2 \beta} = \frac{	an \beta}{2 + 3 	an^2 \beta}$.Since $3 	an \beta = 2 	an \alpha$,we substitute $3 	an \beta$ with $2 	an \alpha$:$\frac{	an \beta}{2 + (2 	an \alpha) 	an \beta} = \frac{	an \alpha - 	an \beta}{1 + 	an \alpha 	an \beta} = 	an (\alpha - \beta)$.

If $\sin (\alpha+\beta)=5 \sin (\alpha-\beta)$,then $\frac{\sin 2 \beta}{5-\cos 2 \beta}=$

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