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प्रश्न
Prove that:
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उत्तर
Consider LHS:
\[ \frac{\sin A + \sin 3A}{\cos A - \cos 3A}\]
\[ = \frac{2\sin \left( \frac{A + 3A}{2} \right) \cos \left( \frac{A - 3A}{2} \right)}{2\sin \left( \frac{A + 3A}{2} \right) \sin \left( \frac{3A - A}{2} \right)} \left\{ \because \sin A + \sin B = 2\sin \left( \frac{A + B}{2} \right) \cos \left( \frac{A - B}{2} \right), and \cos A - \cos B = 2\sin \left( \frac{A + B}{2} \right) cos \left( \frac{B - A}{2} \right) \right\}\]
\[ = \frac{\sin 2A \cos \left( - A \right)}{\sin 2A \sin A}\]
\[ = \frac{\sin 2A \cos A}{\sin 2A \sin A}\]
\[ = \cot A\]
= RHS
Hence, LHS = RHS .
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