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Question
In triangle ABC, prove the following:
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Solution
Let
\[\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k\] ...(1)
Then,
Consider the LHS of the equation
\[LHS = \frac{a + b}{c}\]
\[ = \frac{k\sin A + k\sin B}{k\sin C} \left( using \left( 1 \right) \right)\]
\[ = \frac{2\sin\left( \frac{A + B}{2} \right)\cos\left( \frac{A - B}{2} \right)}{2\sin\frac{C}{2}\cos\frac{C}{2}}\]
\[ = \frac{\sin\left( \frac{A + B}{2} \right)\cos\left( \frac{A - B}{2} \right)}{\sin\frac{C}{2}\cos\left( \frac{\pi - \left( A + B \right)}{2} \right)} \left( \because A + B + C = \pi \right)\]
\[ = \frac{\sin\left( \frac{A + B}{2} \right)\cos\left( \frac{A - B}{2} \right)}{\sin\frac{C}{2}\sin\left( \frac{A + B}{2} \right)}\]
\[ = \frac{\cos\left( \frac{A - B}{2} \right)}{\sin\frac{C}{2}} = RHS \]
\[\text{ Hence proved } .\]
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