Assertion(A): if $x y + y z + z x = 1,$ then $\sum \frac{x}{(1 + x^{2})} = \frac{2}{\sqrt{\prod (1 + x^{2})}}$
Reason (R) : In a $Δ A B C$ $\sin 2 A + \sin 2 B - \sin 2 C = 4 \cos A \cos B \sin C$

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Both A and R are individually true but R is not the correct explanation of A

A is false but R is true

A is true but R is false

Both A and R are individually true and R is the correct explanation of A

answer is B.

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Detailed Solution

Let $x = \cot A, y = \cot B, z = \cot C$

$∴ \cot A \cot B + \cot B \cot C + \cot C \cot A = 1$
$∴ A + B + C = 180^{o}$
$\sum \frac{x}{(1 + x^{2})} = \sum \frac{\cot A}{(1 + \cot^{2} A)} = \frac{1}{2} \sum \frac{2 \tan A}{(1 + \tan^{2} A)}$

$= \frac{1}{2} \sum \sin 2 A$
$= \frac{1}{2} (\sin 2 A + \sin 2 B + \sin 2 C)$
$= \frac{1}{2} (4 \sin A \sin B \sin C)$
$= 2 \sin A \sin B \sin C$
$= \frac{2}{\sqrt{(1 + \cot^{2} A) (1 + \cot^{2} B) (1 + \cot^{2} C)}}$
$= \frac{2}{\sqrt{(1 + x^{2}) (1 + y^{2}) (1 + x^{2})}} = \frac{2}{\sqrt{\prod (1 + x^{2})}}$
and $\sin 2 A + \sin 2 B - \sin 2 C$
$= 2 \sin (A + B) \cos (A - B) - 2 \sin C \cos C$
$= 2 \sin C (\cos (A - B) - \cos C)$
$= 2 \sin C (\cos (A - B) + \cos (A + B))$
$= 2 \sin C (2 \cos A \cos B)$
$= 4 \cos A \cos B \sin C$