If in \(\triangle \mathrm{ABC}\), with usual notations, the angles are in A.P., then \(\frac{a}{c} \sin 2 C+\frac{c}{a} \sin 2 A=\)

Angles \(A, B, C\) of \(\triangle A B C\) are in A.P.
\(\therefore \angle B=60^{\circ} \angle \mathrm{A}=+\angle \mathrm{C}=120^{\circ} \)
\( \text { Also } \frac{\mathrm{a}}{\sin \mathrm{A}}=\frac{\mathrm{b}}{\sin \mathrm{B}}=\frac{\mathrm{c}}{\sin \mathrm{C}} \)
\( \therefore \mathrm{a}=\mathrm{k} \sin \mathrm{A} \ \mathrm{c}=\mathrm{k} \sin \mathrm{C} \)
\( \frac{\mathrm{a}}{\mathrm{c}} \sin 2 \mathrm{C}+\frac{\mathrm{c}}{\mathrm{a}} \sin 2 \mathrm{~A} \)
\( =\frac{\mathrm{k} \sin \mathrm{A}}{\mathrm{k} \sin \mathrm{C}}(2 \sin \mathrm{C} \cos \mathrm{C})+\frac{\mathrm{k} \sin \mathrm{C}}{\mathrm{k} \sin \mathrm{A}}(2 \sin \mathrm{A} \cos \mathrm{A}) \)
\( =2 \sin \mathrm{A} \cos \mathrm{C}+2 \cos \mathrm{A} \sin \mathrm{C}=2(\sin \mathrm{A} \cos \mathrm{C}\) \(+\cos \mathrm{A} \sin \mathrm{C}) \)
\( =2 \sin (\mathrm{A}+\mathrm{C})=2 \sin \left(120^{\circ}\right)=2\left(\frac{\sqrt{3}}{2}\right)=\sqrt{3} \)