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


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