Question:
Prove the following trigonometric identities.
$\sin ^{2} A \cos ^{2} B-\cos ^{2} A \sin ^{2} B=\sin ^{2} A-\sin ^{2} B$
Solution:
We know that, $\sin ^{2} A+\cos ^{2} A=1$
So have,
$\sin ^{2} A \cos ^{2} B-\cos ^{2} A \sin ^{2} B=\sin ^{2} A\left(1-\sin ^{2} B\right)-\left(1-\sin ^{2} A\right) \sin ^{2} B$
$=\sin ^{2} A-\sin ^{2} A \sin ^{2} B-\sin ^{2} B+\sin ^{2} A \sin ^{2} B$
$=\sin ^{2} A-\sin ^{2} B$
Hence proved.
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