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If M Sin θ = N Sin ( θ + 2 α ) , Prove that Tan ( θ + α ) Cot α = M + N M − N - Mathematics

Sum

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:
\[m \sin\theta = n \sin\left( \theta + 2\alpha \right)\] 

\[\Rightarrow \frac{m}{n} = \frac{\sin\left( \theta + 2\alpha \right)}{\sin\theta}\]
Applying componendo and dividendo, we get

\[\frac{m + n}{m - n} = \frac{\sin\left( \theta + 2\alpha \right) + \sin\theta}{\sin\left( \theta + 2\alpha \right) - \sin\theta}\]

\[ \Rightarrow \frac{m + n}{m - n} = \frac{2\sin\left( \frac{\theta + 2\alpha + \theta}{2} \right)\cos\left( \frac{\theta + 2\alpha - \theta}{2} \right)}{2\sin\left( \frac{\theta + 2\alpha - \theta}{2} \right)\cos\left( \frac{\theta + 2\alpha + \theta}{2} \right)}\]

\[ \Rightarrow \frac{m + n}{m - n} = \frac{\sin\left( \theta + \alpha \right) \cos\alpha}{\sin\alpha \cos\left( \theta + \alpha \right)}\]

\[ \Rightarrow \frac{m + n}{m - n} = \tan\left( \theta + \alpha \right) \cot\alpha\]

\[\therefore \tan\left( \theta + \alpha \right) \cot\alpha = \frac{m + n}{m - n}\]
Concept: Transformation Formulae
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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 8 Transformation formulae
Exercise 8.2 | Q 19 | Page 19
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