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).
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