Question
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
A.
$$\frac{1}{2}$$
B.
$$\frac{{\sqrt 3 }}{2}$$
C.
1
D.
$$\sqrt 3 $$
Answer :
$$\sqrt 3 $$
Solution :
$$\eqalign{
& \because \,\,A,B,C\,\,{\text{are in A}}{\text{.P}}{\text{.}} \cr
& \therefore \,\,B = {60^ \circ }\,\,{\text{then}}\frac{a}{c}\sin 2C + \frac{c}{a}\sin 2A \cr
& = \frac{{K\sin A}}{{K\sin C}}.2\sin C\cos C + \frac{{K\sin C}}{{K\sin A}},\sin A\cos A \cr
& = 2\left( {\sin A\cos C + \cos A\sin C} \right) \cr
& = 2\sin \left( {A + C} \right) \cr
& = 2\sin B \cr
& = 2 \times \frac{{\sqrt 3 }}{2} \cr
& = \sqrt 3 \cr} $$