sec4 A − sec2 A is equal to

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).

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