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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:
Consider
\[LHS = \frac{c}{a - b}\]
\[ = \frac{k\sin C}{k\left( \sin A - \sin B \right)} \left( \text{ using } \left( 1 \right) \right)\]
\[ = \frac{2\sin\frac{C}{2}\cos\frac{C}{2}}{2\sin\left( \frac{A - B}{2} \right)\cos\left( \frac{A + B}{2} \right)}\]
\[ = \frac{\sin\left( \frac{\pi - \left( A + B \right)}{2} \right)\cos\frac{C}{2}}{\sin\left( \frac{A - B}{2} \right)\cos\left( \frac{A + B}{2} \right)} \left( \because A + B + C = \pi \right)\]
\[ = \frac{\cos\frac{C}{2}\cos\left( \frac{A + B}{2} \right)}{\sin\left( \frac{A - B}{2} \right)\cos\left( \frac{A + B}{2} \right)}\]
\[ = \frac{\cos\frac{C}{2}}{\sin\left( \frac{A - B}{2} \right)} . . . \left( 2 \right)\]
\[RHS = \frac{\tan\frac{A}{2} + \tan\frac{B}{2}}{\tan\frac{A}{2} - \tan\frac{B}{2}}\]
\[ = \frac{\frac{\sin\frac{A}{2}}{\cos\frac{A}{2}} + \frac{\sin\frac{B}{2}}{\cos\frac{B}{2}}}{\frac{\sin\frac{A}{2}}{\cos\frac{A}{2}} - \frac{\sin\frac{B}{2}}{\cos\frac{B}{2}}}\]
\[ = \frac{\sin\frac{A}{2}\cos\frac{B}{2} + \sin\frac{B}{2}\cos\frac{A}{2}}{\sin\frac{A}{2}\cos\frac{B}{2} - \sin\frac{B}{2}\cos\frac{A}{2}}\]
\[ = \frac{\sin\left( \frac{A + B}{2} \right)}{\sin\left( \frac{A - B}{2} \right)}\]
\[ = \frac{\sin\left( \frac{\pi - C}{2} \right)}{\sin\left( \frac{A - B}{2} \right)}\]
\[ = \frac{\cos\frac{C}{2}}{\sin\left( \frac{A - B}{2} \right)} \]
\[ = LHS \left( \text{ from } \left( 2 \right) \right)\]
\[\text{ Hence proved } .\]
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