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

As the angles A, B, C are in arithmetic progressions,

2B = A + C

Also, as

A + B + C = 180°, 

we get 3B = 180° or B = 60°

By the law of sines

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

Thus,

$\begin{array}{l}\frac{a}{c}\sin 2C+\frac{c}{a}\sin 2A\\ =\frac{\sin A}{\sin C}(2\sin C\cos C)+\frac{\sin C}{\sin A}(2\sin A\cos A)\\ =2(\sin A\cos C+\cos A\sin C)\\ =2\sin (A+C)=2\sin (\pi -B)\\ =2\sin B=2\sin {60}^{\circ }=2\left(\frac{\sqrt{3}}{2}\right)=\sqrt{3}\end{array}$