If A + B + C = pi and sin left( A + frac{C}{2 | vedclass

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(A) Given $\sin \left( A + \frac{C}{2} \right) = k \sin \frac{C}{2}$.Applying Componendo and Dividendo:$\frac{\sin (A + C/2) + \sin (C/2)}{\sin (A + C

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$\frac{k - 1}{k + 1}$

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(A) Given $\sin \left( A + \frac{C}{2} ight) = k \sin \frac{C}{2}$.Applying Componendo and Dividendo:$\frac{\sin (A + C/2) + \sin (C/2)}{\sin (A + C/2) - \sin (C/2)} = \frac{k + 1}{k - 1}$.Using the sum-to-product formulas $\sin x + \sin y = 2 \sin \frac{x+y}{2} \cos \frac{x-y}{2}$ and $\sin x - \sin y = 2 \cos \frac{x+y}{2} \sin \frac{x-y}{2}$:$\frac{2 \sin \frac{A+C}{2} \cos \frac{A}{2}}{2 \cos \frac{A+C}{2} \sin \frac{A}{2}} = \frac{k + 1}{k - 1}$.$	an \left( \frac{A+C}{2} ight) \cot \frac{A}{2} = \frac{k + 1}{k - 1}$.Since $A + B + C = \pi$,we have $\frac{A+C}{2} = \frac{\pi - B}{2} = \frac{\pi}{2} - \frac{B}{2}$.Thus,$	an \left( \frac{\pi}{2} - \frac{B}{2} ight) = \cot \frac{B}{2}$.Substituting this into the equation:$\cot \frac{B}{2} \cot \frac{A}{2} = \frac{k + 1}{k - 1}$.Therefore,$	an \frac{A}{2} 	an \frac{B}{2} = \frac{k - 1}{k + 1}$.

If $A + B + C = \pi$ and $\sin \left( A + \frac{C}{2} \right) = k \sin \frac{C}{2}$,then $\tan \frac{A}{2} \tan \frac{B}{2} = $

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