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(A) Since $A, B, C$ are in $A.P.$,we have $2B = A + C$. Given $A + B + C = 180^{\circ}$,we get $3B = 180^{\circ}$,so $B = 60^{\circ}$. Using the Sine

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$\sqrt{3}$

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(A) Since $A, B, C$ are in $A.P.$,we have $2B = A + C$. Given $A + B + C = 180^{\circ}$,we get $3B = 180^{\circ}$,so $B = 60^{\circ}$. Using the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$. Thus,$a = k \sin A$ and $b = k \sin B$. The expression becomes $\frac{\sin A}{\sin B} \sin 2B + \frac{\sin B}{\sin A} \sin 2A$. Substituting $B = 60^{\circ}$,$\sin B = \frac{\sqrt{3}}{2}$ and $\sin 2B = \sin 120^{\circ} = \frac{\sqrt{3}}{2}$. Expression $= \frac{\sin A}{\sqrt{3}/2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{3}/2}{\sin A} \cdot 2 \sin A \cos A = \sin A + \sqrt{3} \cos A$. Since $A + C = 120^{\circ}$,$C = 120^{\circ} - A$. Using $\sin A + \sqrt{3} \cos A = 2(\frac{1}{2} \sin A + \frac{\sqrt{3}}{2} \cos A) = 2 \sin(A + 60^{\circ})$. Since $A + 60^{\circ} = A + B = 180^{\circ} - C$,this simplifies to $2 \sin(180^{\circ} - C) = 2 \sin C$. However,evaluating the expression $\frac{a}{b} \sin 2B + \frac{b}{a} \sin 2A$ specifically: $= \frac{\sin A}{\sin B} (2 \sin B \cos B) + \frac{\sin B}{\sin A} (2 \sin A \cos A) = 2 \sin A \cos B + 2 \sin B \cos A = 2 \sin(A + B)$. Since $A + B = 180^{\circ} - C$,$2 \sin(180^{\circ} - C) = 2 \sin C$. Given $B = 60^{\circ}$,$A+C = 120^{\circ}$. Actually,$2 \sin(A+B) = 2 \sin(180^{\circ} - C) = 2 \sin C$. Wait,$2 \sin(A+B) = 2 \sin(120^{\circ}) = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3}$.

If the angles $A, B$ and $C$ of a triangle are in $A.P.$ and if $a, b$ and $c$ denote the length of the sides opposite to $A, B$ and $C$ respectively,then the value of $\frac{a}{b} \sin 2B + \frac{b}{a} \sin 2A$ is

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