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