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If in $\triangle \mathrm{ABC}$, with usual notations, the angles are in A.P., then $\frac{a}{c} \sin 2 C+\frac{c}{a} \sin 2 A=$
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The correct answer is:
$\sqrt{3}$
Angles $A, B, C$ of $\triangle A B C$ are in A.P.
$$
\begin{aligned}
& \therefore \angle B=60^{\circ} \angle \mathrm{A}=+\angle \mathrm{C}=120^{\circ} \\
& \text { Also } \frac{\mathrm{a}}{\sin \mathrm{A}}=\frac{\mathrm{b}}{\sin \mathrm{B}}=\frac{\mathrm{c}}{\sin \mathrm{C}} \\
& \therefore \mathrm{a}=\mathrm{k} \sin \mathrm{A} \& \mathrm{c}=\mathrm{k} \sin \mathrm{C} \\
& \frac{\mathrm{a}}{\mathrm{c}} \sin 2 \mathrm{C}+\frac{\mathrm{c}}{\mathrm{a}} \sin 2 \mathrm{~A} \\
& =\frac{\mathrm{k} \sin \mathrm{A}}{\mathrm{k} \sin \mathrm{C}}(2 \sin \mathrm{C} \cos \mathrm{C})+\frac{\mathrm{k} \sin \mathrm{C}}{\mathrm{k} \sin \mathrm{A}}(2 \sin \mathrm{A} \cos \mathrm{A}) \\
& =2 \sin \mathrm{A} \cos \mathrm{C}+2 \cos \mathrm{A} \sin \mathrm{C}=2(\sin \mathrm{A} \cos \mathrm{C}+\cos \mathrm{A} \sin \mathrm{C}) \\
& =2 \sin (\mathrm{A}+\mathrm{C})=2 \sin \left(120^{\circ}\right)=2\left(\frac{\sqrt{3}}{2}\right)=\sqrt{3}
\end{aligned}
$$
$$
\begin{aligned}
& \therefore \angle B=60^{\circ} \angle \mathrm{A}=+\angle \mathrm{C}=120^{\circ} \\
& \text { Also } \frac{\mathrm{a}}{\sin \mathrm{A}}=\frac{\mathrm{b}}{\sin \mathrm{B}}=\frac{\mathrm{c}}{\sin \mathrm{C}} \\
& \therefore \mathrm{a}=\mathrm{k} \sin \mathrm{A} \& \mathrm{c}=\mathrm{k} \sin \mathrm{C} \\
& \frac{\mathrm{a}}{\mathrm{c}} \sin 2 \mathrm{C}+\frac{\mathrm{c}}{\mathrm{a}} \sin 2 \mathrm{~A} \\
& =\frac{\mathrm{k} \sin \mathrm{A}}{\mathrm{k} \sin \mathrm{C}}(2 \sin \mathrm{C} \cos \mathrm{C})+\frac{\mathrm{k} \sin \mathrm{C}}{\mathrm{k} \sin \mathrm{A}}(2 \sin \mathrm{A} \cos \mathrm{A}) \\
& =2 \sin \mathrm{A} \cos \mathrm{C}+2 \cos \mathrm{A} \sin \mathrm{C}=2(\sin \mathrm{A} \cos \mathrm{C}+\cos \mathrm{A} \sin \mathrm{C}) \\
& =2 \sin (\mathrm{A}+\mathrm{C})=2 \sin \left(120^{\circ}\right)=2\left(\frac{\sqrt{3}}{2}\right)=\sqrt{3}
\end{aligned}
$$
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