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Question
If tan A = x tan B, prove that
\[\frac{\sin \left( A - B \right)}{\sin \left( A + B \right)} = \frac{x - 1}{x + 1}\]
Solution
\[\text{ LHS }= \frac{\sin(A - B)}{\sin(A + B)} \]
\[ = \frac{\sin A \cos B - cos A \sin B}{\sin A \cos B + \cos A \sin B}\]
\[\text{ Dividing numerator and denominator by }\cos A \cos B: \]
\[ \frac{\tan A - \tan B}{\tan A + \tan B}\]
\[ = \frac{x\tan B - \tan B}{x\tan B + \tan B} (\text{ Since }\tan A = x \tan B )\]
\[ = \frac{\tan B \left( x - 1 \right)}{\tan B \left( x + 1 \right)}\]
\[ = \frac{x - 1}{x + 1}\]
= RHS
Hence proved.
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