Question

sin^2a×cos^2b+ cos^2a×sin^2b+sin^2a×sin^2b+cos^2a×cos^2b​

Answer 1

Answer:

The given equation is:

sin^2Acos^2B-cos^2Asin^2B=sin^2A-sin^2Bsin

2

Acos

2

B−cos

2

Asin

2

B=sin

2

A−sin

2

B

Taking the left hand side of the above equation, we get

sin^2Acos^2B-cos^2Asin^2Bsin

2

Acos

2

B−cos

2

Asin

2

B

Substituting cos^2B=1-sin^2Bcos

2

B=1−sin

2

B , we get

=sin^2A(1-sin^2B)-(1-sin^2A)sin^2Bsin

2

A(1−sin

2

B)−(1−sin

2

A)sin

2

B

=sin^2A-sin^2Asin^2B-sin^2B+sin^2Asin^2Bsin

2

A−sin

2

Asin

2

B−sin

2

B+sin

2

Asin

2

B

=sin^2A-sin^2Bsin

2

A−sin

2

B

=RHS

Hence proved.