i. Take L.H.S

sin A sin (B – C) + sin B sin (C – A) + sin C sin (A – B)

Multiplying & Dividing by 2:

{∵ 2 sin A sin B = cos (A – B) – cos (A + B)}

= 0

= R.H.S

Hence Proved

ii. sin (B – C) cos (A – D) + sin (C – A) cos(B – D) + sin(A – B) cos(C – D) = 0

Answer:

Take L.H.S

sin (B – C) cos(A – D) + sin (C – A) cos(B – D) + sin(A – B) cos(C – D)

Multiplying & Dividing by 2:

{∵ 2 sin A cos B = sin (A + B) + sin (A – B)}

= 0

= R.H.S

Hence Proved