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Question
In triangle ABC, prove the following:
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Solution
\[\text{ Assume }\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k\]
Consider the LHS of the equation
\[LHS = \frac{a - b}{a + b}\]
\[ = \frac{k\left( \sin A - \sin B \right)}{k\left( \sin A + \sin B \right)}\]
\[\because $\sin A - \sin B = 2\sin\left( \frac{A - B}{2} \right)\cos\left( \frac{A + B}{2} \right), \sin A + \sin B = 2\sin\left( \frac{A + B}{2} \right)\cos\left( \frac{A - B}{2} \right)$\]
\[\therefore LHS = \frac{2\sin\left( \frac{A - B}{2} \right)\cos\left( \frac{A + B}{2} \right)}{2\sin\left( \frac{A + B}{2} \right)\cos\left( \frac{A - B}{2} \right)} $$\]
\[ = \frac{\tan\left( \frac{A - B}{2} \right)}{\tan\left( \frac{A + B}{2} \right)} = RHS \]
\[\text{ Hence proved } .\]
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