JEE Advance - Mathematics (2010 - Paper 1 Offline - No. 12)
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 $${a \over c}\sin 2C + {c \over a}\sin 2A$$ is
$${1 \over 2}$$
$${{\sqrt 3 } \over 2}$$
$$1$$
$${\sqrt 3 }$$
Explanation
Since, A, B, C are in AP
$$\Rightarrow$$ 2B = A + C i.e., $$\angle$$B = 60$$^\circ$$
$$\therefore$$ $${a \over c}$$(2 sin C cos C) + $${c \over a}$$ (2 sin A cos A)
= 2k (a cos C + c cos A)
[using, $${a \over {\sin A}} = {b \over {\sin B}} = {c \over {\sin C}} = {1 \over k}$$]
= 2k (b)
= 2 sin B
[using, b = a cos C + c cos A]
= $$\sqrt3$$
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