MCQ
$\sec ^4 A-\sec ^2 A$ is equal to
- A$\tan ^2 A-\tan ^4 A$
- B$\tan ^4 A-\tan ^2 A$
- C$\tan ^2 A+\tan ^3 A$
- ✓$\tan ^4 A+\tan ^2 A$
✓
Answer
Correct option: D.
$\tan ^4 A+\tan ^2 A$
We have, $\sec ^4 A-\sec ^2 A$
$=\sec ^2 A\left(\sec ^2 A-1\right)$
$=\left(1+\tan ^2 A\right) \tan ^2 A$
$=\tan ^2 A+\tan ^4 A$
$=\tan ^4 A+\tan ^2 A$
$=\sec ^2 A\left(\sec ^2 A-1\right)$
$=\left(1+\tan ^2 A\right) \tan ^2 A$
$=\tan ^2 A+\tan ^4 A$
$=\tan ^4 A+\tan ^2 A$
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