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Question
Prove that:
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Solution
Consider LHS:
\[ \frac{\sin A + \sin 3A + \sin 5A}{\cos A + \cos 3A + \cos 5A}\]
\[ = \frac{\sin A + \sin 5A + \sin 3A}{\cos A + \cos 5A + \cos 3A}\]
\[ = \frac{2\sin \left( \frac{A + 5A}{2} \right) \cos \left( \frac{A - 5A}{2} \right) + \sin 3A}{2\cos \left( \frac{A + 5A}{2} \right) \cos \left( \frac{A - 5A}{2} \right) + \cos 3A}\]
\[ \]
\[ = \frac{2 \sin 3A \cos \left( - 2A \right) + \sin 3A}{2\cos 3A \cos \left( - 2A \right) + \cos 3A}\]
\[ = \frac{2\sin 3A \cos 2A + \sin 3A}{2\cos 3A \cos 2A + \cos 3A}\]
\[ = \frac{\sin 3A \left[ 2\cos 2A + 1 \right]}{\cos 3A \left[ 2\cos 2A + 1 \right]}\]
\[ = \tan 3A\]
= RHS
Hence, RHS = LHS .
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