Question

Prove the following trigonometric identities. sin²Acos²B-cos²Asin²B=sin²A-sin²B

Answer 1

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..

HOPE THIS ANSWER WILL HELP YOU…

Answer 2

Answer:

Hey mate plzz refer to the attachment