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