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