9 sec2 A − 9 tan2 A is equal to
Question:

$9 \sec ^{2} A-9 \tan ^{2} A$ is equal to

(a) 1
(b) 9
(c) 8
(d) 0

Solution:

Given:

$9 \sec ^{2} A-9 \tan ^{2} A$

$=9\left(\sec ^{2} A-\tan ^{2} A\right)$

We know that, $\sec ^{2} A-\tan ^{2} A=1$

Therefore, $9 \sec ^{2} A-9 \tan ^{2} A=9$

Hence, the correct option is (b).