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