In Delta ABC,frac{cos frac{1}{2}(B - C)}{sin | vedclass

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(B) We know that in $\Delta ABC$,$A + B + C = 180^{\circ}$,so $\frac{A}{2} = 90^{\circ} - \frac{B + C}{2}$.Thus,$\sin \frac{A}{2} = \sin(90^{\circ} -

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$\frac{b + c}{a}$

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(B) We know that in $\Delta ABC$,$A + B + C = 180^{\circ}$,so $\frac{A}{2} = 90^{\circ} - \frac{B + C}{2}$.Thus,$\sin \frac{A}{2} = \sin(90^{\circ} - \frac{B + C}{2}) = \cos \frac{B + C}{2}$.Substituting this into the expression:$\frac{\cos \frac{B - C}{2}}{\sin \frac{A}{2}} = \frac{\cos \frac{B - C}{2}}{\cos \frac{B + C}{2}}$.Using the formula $2 \sin X \cos Y = \sin(X + Y) + \sin(X - Y)$,we multiply numerator and denominator by $2 \sin \frac{A}{2}$ (or $2 \cos \frac{B + C}{2}$):$= \frac{2 \sin \frac{A}{2} \cos \frac{B - C}{2}}{2 \sin \frac{A}{2} \cos \frac{B + C}{2}} = \frac{\sin(\frac{A}{2} + \frac{B - C}{2}) + \sin(\frac{A}{2} - \frac{B - C}{2})}{\sin A}$.Since $\frac{A}{2} = 90^{\circ} - \frac{B + C}{2}$,this simplifies to $\frac{\sin B + \sin C}{\sin A}$.By the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$,so $\sin A = \frac{a}{k}, \sin B = \frac{b}{k}, \sin C = \frac{c}{k}$.Therefore,$\frac{\sin B + \sin C}{\sin A} = \frac{b + c}{a}$.

In $\Delta ABC$,$\frac{\cos \frac{1}{2}(B - C)}{\sin \frac{1}{2}A} = $

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