$sin (B + C - A) + sin (C + A - B) + sin (A + B - C) =$

4 sin A sin B sin C

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4 cos A cos B cos C

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4 sin (A / 2) sin (B / 2) sin (C / 2)

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N o n e o f t h e s e

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Solution

The correct option is A $4 sin A sin B sin C$
According to question,
$\begin{matrix} sin (B + C - A) + sin (C + A - B) + sin (A + B - C) w e k n o w,,, Δ A + B + C = 180^{0} = sin (π - A - A) + sin (π - B - B) + sin (π - C - C) = sin (π - 2 A) + sin (π - 2 B) + sin (π - 2 C) = s i n 2 A + s i n 2 B + s i n 2 C = 2 s i n (A + B) c o s (A - B) + s i n 2 C = 2 s i n (π - C) c o s (A - B) + s i n 2 C = 2 s i n C c o s (A - B) + s i n 2 C = 2 s i n C c o s (A - B) + 2 s i n C c o s C = 2 s i n C [cos (A - B) + cos C] = 2 sin C [(A - B) + cos (π - (A + B))] = 2 sin C [cos (A - B) + (- 1) cos (A + B)] = 2 sin C [C O S (A - B) - cos (A = B)] = 2 sin C (- 1) 2 sin \frac{(A - B + A + B)}{2} \times sin \frac{(A - B - A - B)}{2} = 2 sin C - 2 sin A sin (- B) [w e k n o w, sin (- θ) = - sin θ] = 4 sin C . s i n A . s i n B ∴ 4 s i n A s i n B s i n C \end{matrix}$
So,that the correct option is A.

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