Question:
$\sec ^{4} A-\sec ^{2} A$ is equal to
(a) $\tan ^{2} \mathrm{~A}-\tan ^{4} \mathrm{~A}$
(b) $\tan ^{4} A-\tan ^{2} A$
(c) $\tan ^{4} A+\tan ^{2} A$
(d) $\tan ^{2} \mathrm{~A}+\tan ^{4} \mathrm{~A}$
Solution:
The given expression is $\sec ^{4} \mathrm{~A}-\sec ^{2} \mathrm{~A}$.
Taking common $\sec ^{2} \mathrm{~A}$ from both the terms, we have
$\sec ^{4} A-\sec ^{2} A$
$=\sec ^{2} \mathrm{~A}\left(\sec ^{2} \mathrm{~A}-1\right)$
$=\left(1+\tan ^{2} \mathrm{~A}\right) \tan ^{2} \mathrm{~A}$
$=\tan ^{2} \mathrm{~A}+\tan ^{4} \mathrm{~A}$
Disclaimer: The options given in (c) and (d) are same by the commutative property of addition.
Therefore, the correct options are (c) or (d).