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Question
If cos A = m cos B, then \[\cot\frac{A + B}{2} \cot\frac{B - A}{2}\]=
Options
- \[\frac{m - 1}{m + 1}\]
- \[\frac{m + 2}{m - 2}\]
- \[\frac{m + 1}{m - 1}\]
None of these
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Solution
Given:
\[\cos A = m\cos B\]
\[ \Rightarrow \frac{\cos A}{\cos B} = \frac{m}{1}\]
\[ \Rightarrow \frac{\cos A + \cos B}{\cos A - \cos B} = \frac{m + 1}{m - 1}\]
\[ \Rightarrow \frac{2\cos\left( \frac{A - B}{2} \right)\cos\left( \frac{A + B}{2} \right)}{- 2\sin\left( \frac{B + A}{2} \right)\sin\left( \frac{B - A}{2} \right)} = \frac{m + 1}{m - 1} \left[ \because \cos A + \cos B = 2\cos\left( \frac{A - B}{2} \right)\cos\left( \frac{A + B}{2} \right) \text{ and }\cos A - \cos B = 2\sin\left( \frac{A + B}{2} \right)\cos\left( \frac{B - A}{2} \right) \right]\]
\[ \Rightarrow \frac{\cos\left( \frac{B - A}{2} \right)\cos\left( \frac{A + B}{2} \right)}{\sin\left( \frac{A + B}{2} \right)\sin\left( \frac{B - A}{2} \right)} = \frac{m + 1}{m - 1} \]
\[ \Rightarrow \cot\left( \frac{A + B}{2} \right)\cot\left( \frac{B - A}{2} \right)=\frac{m + 1}{m - 1}\]
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