Question
Prove the trigonometric identity: $\tan ^2 A \sec ^2 B-\sec ^2 A \tan ^2 B=\tan ^2 A-\tan ^2 B$
✓
Answer
$\text { LHS }=\tan ^2 A \sec ^2 B-\sec ^2 A \tan ^2 B$
$=\tan ^2 A\left(1+\tan ^2 B\right)-\left(1+\tan ^2 A\right) \tan ^2 B\left [\because 1+\tan ^2 \theta=\sec ^2 \theta\right]$
$=\tan ^2 A+\tan ^2 A \tan ^2 B-\tan ^2 B-\tan ^2 A \tan ^2 B$
$=\tan ^2 A-\tan ^2 B$
$=\text { RHS }$
Hence proved.
$=\tan ^2 A\left(1+\tan ^2 B\right)-\left(1+\tan ^2 A\right) \tan ^2 B\left [\because 1+\tan ^2 \theta=\sec ^2 \theta\right]$
$=\tan ^2 A+\tan ^2 A \tan ^2 B-\tan ^2 B-\tan ^2 A \tan ^2 B$
$=\tan ^2 A-\tan ^2 B$
$=\text { RHS }$
Hence proved.
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