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