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Question
Prove that:
Solution
Consider LHS:
\[ \frac{\cos A + cos B}{\cos B - \cos A}\]
\[ = \frac{2\cos \left( \frac{A - B}{2} \right) \cos \left( \frac{A + B}{2} \right)}{2\sin \left( \frac{A + B}{2} \right) \sin \left( \frac{A - B}{2} \right)} \left[ \because \cos A + \cos B = 2\cos \left( \frac{A - B}{2} \right) \cos \left( \frac{A + B}{2} \right) and \cos A - \cos B = 2\sin \left( \frac{A + B}{2} \right) cos \left( \frac{B - A}{2} \right) \right]\]
\[ = \frac{\cos \left( \frac{A - B}{2} \right) \cos \left( \frac{A + B}{2} \right)}{\sin \left( \frac{A + B}{2} \right) sin \left( \frac{A - B}{2} \right)}\]
\[ = \cot\left( \frac{A + B}{2} \right)\cot\left( \frac{A - B}{2} \right)\]
=RHS
Hence, LHS = RHS.
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