Question

prove that:
sin A . sin (B-C) + sin B. sin (C - A) + sin C sin (A - B) = 0​

Answer 1

Answer:

It have to be given, In triangle ABC,

As sinA = sin[180 - (B + C)] = sin (B + C)

similarly, sinB = sin(A + C)

similarly, sinC = sin(A + B)

Consider

sinA.sin(B- C) = sin(B + C). sin(B- C) = sin^2B - sin^2C

sinB.sin(C - A) = sin(A + C). sin(C - A) = sin^2C - sin^2A

sinC.sin(A - B) = sin(B + A). sin(A - B) = sin^2A - sin^2B

On adding the above three equations, we get

sin A . sin (B-C) + sin B. sin (C - A) + sin C sin (A - B) = 0