Question

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

Answer 1

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Answer 2

Answer:

Step-by-step explanation:

The given equation is:

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

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

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

Substituting $cos^2B=1-sin^2B$, we get

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

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

=$sin^2A-sin^2B$

=RHS

Hence proved.