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

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Aicharya · Accepted Answer

$\begin{aligned} & 	ext { } \sin (\alpha+\beta)=5 \sin (\alpha-\beta) \ & \Rightarrow \sin \alpha \cdot \cos \beta+\cos \alpha \cdot \sin \beta\end{aligned}$ $\begin{aligned} & -5 \sin \alpha \cos \beta-5 \cos \alpha \cdot \sin \beta \ & \Rightarrow 6 \cos \alpha \cdot \sin \beta=4 \sin \alpha \cos \beta\end{aligned}$ $\Rightarrow 3 	an \beta=2 	an \alpha$ $ \begin{aligned} & 	ext { Now } \frac{\sin 2 \beta}{5-\cos 2 \beta}=\frac{\left(\frac{2 	an \beta}{1+	an ^2 \beta}ight)}{5-\left(\frac{1-	an ^2 \beta}{1+	an ^2 \beta}ight)} \ & \left\{\begin{array}{r} \because 	ext { (i) } \sin 2 \beta=\frac{2 	an \beta}{1+	an ^2 \beta} \ 	ext { (ii) } \cos 2 \beta=\frac{1-	an ^2 \beta}{1+	an ^2 \beta} \end{array}ight. \ & \Rightarrow \frac{\sin 2 \beta}{5-\cos 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{2[3 	an \beta-2 	an \beta]}{4\left[1+\frac{3}{2} 	an ^2 \betaight]} \ & =\frac{[2 	an \alpha-2 	an \beta]}{2\left[1+\left(\frac{3 	an \beta}{2}ight) \cdot 	an \betaight]} \ & =\frac{	an \alpha-	an \beta}{1+\left(\frac{2 	an \alpha}{2}ight) \cdot 	an \beta} \ & 	ext { (from eq. (i) } \ & =\frac{	an \alpha-	an \beta}{1+	an \alpha \cdot 	an \beta}=	an (\alpha-\beta) \ & \end{aligned} $ (from eq. (i)

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

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