In triangle ABC, $sin (B + C - A) + sin (C + A - B) + sin (B + A - C) = k sin A sin B sin C$
Find $k$ .

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Solution

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

$= sin (π - 2 A) + sin (π - 2 B) + sin (π - 2 C)$ ................... As $A + B + C = π$

$= sin 2 A + sin 2 B + sin 2 C = 2 sin (A + B) cos (A - B) + 2 sin C cos C$

$= 2 sin C cos (A - B) + 2 sin C cos C = 2 sin C (cos (A - B) - cos (A + B)) = 4 sin A sin B sin C$

$k = 4$
Ans: 4

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