If the angles$A, B, C$of a triangle are in $A.P.,$ then $\frac{a}{c}\sin 2C+\frac{c}{a}\sin 2A=$

Solution:

$A,B,C$ are in  A.P Then$\Rightarrow B={60}^{\circ }$

$\begin{array}{l}\therefore \frac{a}{c}\sin 2C+\frac{c}{a}\sin 2A\\ =\frac{\sin A}{\sin C}\cdot 2\sin C\cos C+\frac{\sin C}{\sin A}\cdot 2\sin A\cos A\end{array}$

$$\begin{gathered} =2(\sin A\cos C+\cos A\sin C) \\ =2\sin (A+C)=2\sin {120}^{\circ }=\sqrt{3} \end{gathered}$$