It should be:
Prove that: sin²B = sin²A + sin²(A - B) - 2*sin(A)*cos(B)*sin(A - B)
ii) Let me prove the same by taking Right side = Left side.
Applying 2sin(A)*cos(B) = sin(A + B) + sin(A - B), Right side is:
= sin²A + sin²(A - B) - sin(A - B){sin(A + B) + sin(A - B)}
= sin²A + sin²(A - B) - sin(A + B)*sin(A - B) - sin²(A - B)
= sin²A - {sin(A + B)*sin(A - B)}
= sin²A - sin²A + sin²B [Since sin(A + B)*sin(A - B) = sin²A - sin²B]
= sin²B = Left side [Proved]
Proof for sin(A + B)*sin(A - B) = sin²A - sin²B:
sin(A + B)*sin(A - B) = {sin(A)*cos(B) + cos(A)*sin(B)}*{sin(A)*cos(B) - cos(A)*sin(B)}
= sin²A*cos²B - cos²A*sin²B [Applying (a + b)(a - b) = a² - b²]
= sin²A(1 - sin²B) - (1 - sin²A)sin²B
= sin²A - sin²A*sin²B - sin²B + sin²A*sin²B
= sin²A - sin²B
