Prove the following trigonometric identities. sin²Acos²B-cos²Asin²B=sin²A-sin²B
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Answer with Step-by-step explanation:
Given : sin²A cos²B – cos²A sin²B = sin²A – sin²B
LHS = sin²A cos²B – cos²A sin²B
= sin²A(1 − sin²B) − (1 − sin²A)(sin²B)
[By using an identity, cos²θ = (1- sin²θ) ]
= sin²A − sin²A sin²B − sin²B + sin²A sin²B
= sin²A − sin²B − sin²A sin²B + sin²A sin²B
= sin²A − sin²B
sin²A cos²B – cos²A sin²B = sin²A – sin²B
L.H.S = R.H.S
Hence Proved..
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