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(D) Given that $A, B, C$ are in an Arithmetic Progression ($A$.$P$.).Since $A + B + C = 180^{\circ}$,we have $3B = 180^{\circ}$,which implies $B = 60^

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$\sqrt{3}$

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(D) Given that $A, B, C$ are in an Arithmetic Progression ($A$.$P$.).Since $A + B + C = 180^{\circ}$,we have $3B = 180^{\circ}$,which implies $B = 60^{\circ}$.By the sine rule,$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} = k$,so $\sin A = ak, \sin B = bk, \sin C = ck$.The expression is $E = \frac{a}{c} \sin 2C + \frac{c}{a} \sin 2A$.$E = \frac{a}{c} (2 \sin C \cos C) + \frac{c}{a} (2 \sin A \cos A)$.Substituting $\sin C = ck$ and $\sin A = ak$:$E = \frac{a}{c} (2 ck \cos C) + \frac{c}{a} (2 ak \cos A) = 2ak \cos C + 2ck \cos A$.$E = 2k (a \cos C + c \cos A)$.Using the projection formula $b = a \cos C + c \cos A$,we get:$E = 2kb = 2 \sin B$.Since $B = 60^{\circ}$,$E = 2 \sin 60^{\circ} = 2 	imes \frac{\sqrt{3}}{2} = \sqrt{3}$.

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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