In a triangle ABC,frac{a-b}{a+b} = | vedclass

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(C) Using the sine rule,we have $a = k \sin A$ and $b = k \sin B$. Substituting these into the expression: $\frac{a-b}{a+b} = \frac{k \sin A - k \sin

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$	an \left(\frac{A-B}{2}ight) 	an \frac{C}{2}$

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(C) Using the sine rule,we have $a = k \sin A$ and $b = k \sin B$. Substituting these into the expression: $\frac{a-b}{a+b} = \frac{k \sin A - k \sin B}{k \sin A + k \sin B} = \frac{\sin A - \sin B}{\sin A + \sin B}$ Using the sum-to-product formulas: $= \frac{2 \cos \left(\frac{A+B}{2}ight) \sin \left(\frac{A-B}{2}ight)}{2 \sin \left(\frac{A+B}{2}ight) \cos \left(\frac{A-B}{2}ight)}$ $= \cot \left(\frac{A+B}{2}ight) 	an \left(\frac{A-B}{2}ight)$ Since $A+B+C = 180^{\circ}$,we have $\frac{A+B}{2} = 90^{\circ} - \frac{C}{2}$. Therefore,$\cot \left(\frac{A+B}{2}ight) = \cot \left(90^{\circ} - \frac{C}{2}ight) = 	an \frac{C}{2}$. Thus,$\frac{a-b}{a+b} = 	an \frac{C}{2} 	an \left(\frac{A-B}{2}ight)$.

In a $\triangle ABC$,$\frac{a-b}{a+b} = $

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