If the angles A, B and C of a triangle are in | vedclass

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(D) Since $A, B, C$ are in arithmetic progression,$2B = A + C$. Given $A + B + C = 180^{\circ}$,we have $3B = 180^{\circ}$,so $B = 60^{\circ}$. Using

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$\sqrt{3}$

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(D) Since $A, B, C$ are in arithmetic progression,$2B = A + C$. Given $A + B + C = 180^{\circ}$,we have $3B = 180^{\circ}$,so $B = 60^{\circ}$. Using the sine rule,$\frac{a}{\sin A} = \frac{c}{\sin C} = 2R$,so $a = 2R \sin A$ and $c = 2R \sin C$. The expression becomes $\frac{\sin A}{\sin C} \sin 2C + \frac{\sin C}{\sin A} \sin 2A$. $= \frac{\sin A}{\sin C} (2 \sin C \cos C) + \frac{\sin C}{\sin A} (2 \sin A \cos A) = 2 \sin A \cos C + 2 \sin C \cos A$. $= 2 \sin(A + C) = 2 \sin(180^{\circ} - B) = 2 \sin B$. Since $B = 60^{\circ}$,the value is $2 \sin 60^{\circ} = 2 	imes \frac{\sqrt{3}}{2} = \sqrt{3}$.

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

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