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Question
Prove that: \[\frac{\sin \left( A + B \right) + \sin \left( A - B \right)}{\cos \left( A + B \right) + \cos \left( A - B \right)} = \tan A\]
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Solution
\[\text{ LHS }= \frac{\sin\left( A + B \right) + \sin\left( A - B \right)}{\cos\left( A + B \right) + \cos\left( A - B \right)}\]
\[ = \frac{\sin A \cos B + \cos A \sin B + \sin A \cos B - \cos A \sin B}{\cos A \cos B - \sin A \sin B + \cos A \cos B + \sin A \sin B}\]
\[ = \frac{2\sin A \cos B}{2\cos A \cos B}\]
\[ = \frac{\sin A}{\cos A}\]
\[ = \tan A\]
= RHS
Hence proved .
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