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Question
If y sin ϕ = x sin (2θ + ϕ), prove that (x + y) cot (θ + ϕ) = (y − x) cot θ.
Solution
Given:
y sin ϕ = x sin (2θ + ϕ)
\[\Rightarrow \frac{y}{x} = \frac{\sin\left( 2\theta + \phi \right)}{\sin\phi}\]
Applying componendo and dividendo:
\[ \Rightarrow \frac{y - x}{y + x} = \frac{\sin\left( 2\theta + \phi \right) - \sin\phi}{\sin\left( 2\theta + \phi \right) + \sin\phi}\]
\[ \Rightarrow \frac{y - x}{y + x} = \frac{2\sin\left( \frac{2\theta + \phi - \phi}{2} \right)\cos\left( \frac{2\theta + \phi + \phi}{2} \right)}{2\sin\left( \frac{2\theta + \phi + \phi}{2} \right)\cos\left( \frac{2\theta + \phi - \phi}{2} \right)}\]
\[ \Rightarrow \frac{y - x}{y + x} = \frac{2\sin \theta \cos\left( \theta + \phi \right)}{2\sin\left( \theta + \phi \right) \cos \theta}\]
\[ \Rightarrow \frac{y - x}{y + x} = \frac{\sin \theta \cos\left( \theta + \phi \right)}{\sin\left( \theta + \phi \right) \cos \theta}\]
\[ \Rightarrow \frac{y - x}{y + x} = \frac{\cot \left( \theta + \phi \right)}{\cot \theta}\]
\[ \Rightarrow \left( y - x \right) cot\theta = \left( y + x \right) cot\left( \theta + \phi \right)\]
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