If $m \sin \theta=n \sin (\theta+2 \alpha)$, prove that $\tan (\theta+\alpha) \cot \alpha=\frac{m+n}{m-n}$. [NCERT EXEMPLAR]
Given: $m \sin \theta=n \sin (\theta+2 \alpha)$
$\Rightarrow \frac{m}{n}=\frac{\sin (\theta+2 \alpha)}{\sin \theta}$
Applying componendo and dividendo, we get
$\frac{m+n}{m-n}=\frac{\sin (\theta+2 \alpha)+\sin \theta}{\sin (\theta+2 \alpha)-\sin \theta}$
$\Rightarrow \frac{m+n}{m-n}=\frac{2 \sin \left(\frac{\theta+2 \alpha+\theta}{2}\right) \cos \left(\frac{\theta+2 \alpha-\theta}{2}\right)}{2 \sin \left(\frac{\theta+2 \alpha-\theta}{2}\right) \cos \left(\frac{\theta+2 \alpha+\theta}{2}\right)}$
$\Rightarrow \frac{m+n}{m-n}=\frac{\sin (\theta+\alpha) \cos \alpha}{\sin \alpha \cos (\theta+\alpha)}$
$\Rightarrow \frac{m+n}{m-n}=\tan (\theta+\alpha) \cot \alpha$
$\therefore \tan (\theta+\alpha) \cot \alpha=\frac{m+n}{m-n}$
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