Aptitude Question ID : 94460

If $\angle A$ and $\angle B$ are complementary to each other, then the value of $\sec ^{2}A+\sec ^{2}B-\sec ^{2}A\cdot \sec ^{2}B$ is :
[A]-1
[B]0
[C]1
[D]2

0
$\angle A +\angle B = 90^{\circ}$
$=> \angle B = 90^{\circ}- \angle A$
$\therefore \sec ^{2}A+\sec ^{2}B-\sec ^{2}A\cdot \sec ^{2}A$
$= \sec ^{2}A+\csc ^{2}A-\sec ^{2}A\cdot \csc ^{2}A$
$= \frac{1}{\cos ^{2}A}+\frac{1}{\sin ^{2}A}-\frac{1}{\sin ^{2}A\cdot \cos ^{2}A}$
$=\frac{\sin ^{2}A+ \cos ^{2}A-1}{\sin ^{2}A\cdot \cos ^{2}A}$
$= \frac{1-1}{\sin ^{2}A\cdot \cos ^{2}A} = 0$
Hence option [B] is correct answer.

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