Prove that:
i.
ii. sin(B– C)cos(A – D) + sin(C – A)cos(B – D) + sin(A – B)cos(C – D) = 0
Take L.H.S.
{sin (-A) = -sin A & cos (-A) = cos A}
= cot C
= R.H.S.
Hence Proved
ii. 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)}
{sin (-A) = -sin A}
= 0
= R.H.S.
Hence Proved