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Question
Prove that:
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Solution
Consider LHS:
\[\frac{\sin 9A - \sin 7A}{\cos 7A - \cos 9A}\]
\[ = \frac{2\sin \left( \frac{9A - 7A}{2} \right) \cos \left( \frac{9A + 7A}{2} \right)}{2\sin \left( \frac{7A + 9A}{2} \right) \sin \left( \frac{9A - 7A}{2} \right)} \left[ \because \sin A - \sin B = 2\sin \left( \frac{A - B}{2} \right) \cos \left( \frac{A + B}{2} \right) and \cos A - \cos B = 2\sin \left( \frac{A + B}{2} \right) \cos \left( \frac{B - A}{2} \right) \right]\]
\[ = \frac{\sin A \cos 8A}{\sin 8A \sin A}\]
\[ = \cot8A\]
= RHS
Hence, LHS = RHS .
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