If the angles A, B, C of a triangle are in A.P., then acsin⁡2C+casin⁡2A=

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

1. A

$\frac{1}{2}$

2. B

$\frac{\sqrt{3}}{2}$

3. C

1

4. D

$\sqrt{3}$

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Solution:

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

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

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