If the angles \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\) of a triangle are in an Arithmetic Progression and if \(\mathrm{a}, \mathrm{b}\) and \(\mathrm{c}\) denote the lengths of the sides opposite to A, B and C respectively, then the value of the expression \(\frac{\mathrm{a}}{\mathrm{c}} \sin 2 \mathrm{C}+\frac{\mathrm{c}}{\mathrm{a}} \sin 2 \mathrm{~A}\) is

A, B, C are in A.P.
\(
\begin{array}{ll}
\therefore \quad & A+C=2 B \\
& \text { Also, } A+B+C=180^{\circ} \\
& \angle B=60^{\circ} \\
& B y \operatorname{sine} \text { rule, } \\
& \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=k \\
\therefore \quad & \sin A=a k, \sin B=b k, \sin C=c k \\
\therefore \quad & \frac{a}{c} \sin 2 C+\frac{c}{a} \sin 2 A \\
& =\frac{a}{c}(2 \sin C \cos C)+\frac{c}{a}(2 \sin A \cos A) \\
& =\frac{a}{c}(2 \operatorname{ck} \cos C)+\frac{c}{a}(2 a k \cos A) \\
& =2 k a \cos C+2 k c \cos A \\
& =2 k(a \cos C+c \cos A)
\end{array}
\)
\(\begin{array}{ll}=2 \mathrm{~kb} & \ldots[\because b=\mathrm{a} \cos \mathrm{C}+\mathrm{c} \cos \mathrm{A}] \\ =2 \sin \mathrm{B} & \\ =2 \times \frac{\sqrt{3}}{2} & \ldots \ldots\left[\because \angle \mathrm{B}=60^{\circ}\right] \\ =\sqrt{3} & \end{array}\)