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Question
In triangle ABC, prove the following:
Solution
Let
\[\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k\] ...(1)
We need to prove:
\[ = \frac{ksinC}{k\left( sinA + sinB \right)} \left( using\left( 1 \right) \right)\]
\[ = \frac{2\sin\frac{C}{2}\cos\frac{C}{2}}{2\sin\left( \frac{A + B}{2} \right)\cos\left( \frac{A - B}{2} \right)}\]
\[ = \frac{\sin\frac{C}{2}\cos\left( \frac{\pi - \left( A + B \right)}{2} \right)}{\sin\left( \frac{A + B}{2} \right)\cos\left( \frac{A - B}{2} \right)} \left( \because A + B + C = \pi \right)\]
\[ = \frac{\sin\frac{C}{2}\sin\left( \frac{A + B}{2} \right)}{\sin\left( \frac{A + B}{2} \right)\cos\left( \frac{A - B}{2} \right)}\]
\[ = \frac{\sin\frac{C}{2}}{\cos\left( \frac{A - B}{2} \right)} . . . \left( 2 \right)\]
\[RHS = \frac{1 - tan\frac{A}{2}tan\frac{B}{2}}{1 + tan\frac{A}{2}tan\frac{B}{2}}\]
\[ = \frac{1 - \frac{sin\frac{A}{2}}{cos\frac{A}{2}}\frac{sin\frac{B}{2}}{cos\frac{B}{2}}}{1 + \frac{sin\frac{A}{2}}{cos\frac{A}{2}}\frac{sin\frac{B}{2}}{cos\frac{B}{2}}}\]
\[ = \frac{cos\frac{A}{2}cos\frac{B}{2} - sin\frac{A}{2}sin\frac{B}{2}}{cos\frac{A}{2}cos\frac{B}{2} + sin\frac{A}{2}sin\frac{B}{2}}\]
\[ = \frac{cos\left( \frac{A + B}{2} \right)}{cos\left( \frac{A - B}{2} \right)}\]
\[ = \frac{cos\left( \frac{\pi - C}{2} \right)}{cos\left( \frac{A - B}{2} \right)} \left( \because A + B + C = \pi \right)\]
\[ = \frac{\sin\frac{C}{2}}{\cos\left( \frac{A - B}{2} \right)} = LHS \left( \text{ from }\left( 2 \right) \right)\]
\[\text{ Hence proved } .\]
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