We have $m \sin \theta=n \sin (\theta+2 \alpha)$
$$
\therefore \quad \frac{\sin (\theta+2 \alpha)}{\sin \theta}=\frac{m}{n}
$$
Using componendo and dividendo, we get
$$
\begin{aligned}
& \frac{\sin (\theta+2 \alpha)+\sin \theta}{\sin (\theta+2 \alpha)-\sin \theta}=\frac{m+n}{m-n} \\
\Rightarrow & \frac{2 \sin \left(\frac{\theta+2 \alpha+\theta}{2}\right) \cdot \cos \left(\frac{\theta+2 \alpha-\theta}{2}\right)}{2 \cos \left(\frac{\theta+2 \alpha+\theta}{2}\right) \cdot \sin \left(\frac{\theta+2 \alpha-\theta}{2}\right)}=\frac{m+n}{m-n} \\
{\left[\because \sin x+\sin y=2 \sin \frac{x+y}{2} \cdot \cos \frac{x-y}{2} \text { and } \sin x-\sin y\right.} \\
\left.=2 \cos \frac{x+y}{2} \sin \cdot \frac{x-y}{2}\right]
\end{aligned}
$$
$$
\begin{aligned}
&\Rightarrow \quad \frac{\sin (\theta+\alpha) \cdot \cos \alpha}{\cos (\theta+\alpha) \cdot \sin \alpha}=\frac{m+n}{m-n} \\
&\Rightarrow \quad \tan (\theta+\alpha) \cdot \cot \alpha=\frac{m+n}{m-n}
\end{aligned}
$$
Hence proved.