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Question
Prove that:
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Solution
Consider LHS:
\[ \frac{\cos 4A + \cos 3A + \cos 2A}{\sin 4A + \sin 3A + \sin 2A}\]
\[ = \frac{\cos 4A + \cos 2A + \cos 3A}{\sin 4A + \sin 2A + \sin 3A}\]
\[ = \frac{2\cos \left( \frac{4A + 2A}{2} \right) \cos \left( \frac{4A - 2A}{2} \right) + \cos 3A}{2\sin \left( \frac{4A + 2A}{2} \right) \cos \left( \frac{4A - 2A}{2} \right) + \sin 3A}\]
\[ \]
\[ = \frac{2\cos 3A \cos A + \cos 3A}{2\sin 3A \cos A + \sin 3A}\]
\[ = \frac{\cos 3A\left[ 2\cos A + 1 \right]}{\sin 3A\left[ 2\cos A + 1 \right]}\]
\[ = \cot 3A\]
= RHS
Hence, RHS = LHS.
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