Aptitude Question ID : 94460
If $latex \angle A$ and $latex \angle B$ are complementary to each other, then the value of $latex \sec ^{2}A+\sec ^{2}B-\sec ^{2}A\cdot \sec ^{2}B$ is :
[A]-1
[B]0
[C]1
[D]2
0
$latex \angle A +\angle B = 90^{\circ}$
$latex => \angle B = 90^{\circ}- \angle A$
$latex \therefore \sec ^{2}A+\sec ^{2}B-\sec ^{2}A\cdot \sec ^{2}A$
$latex = \sec ^{2}A+\csc ^{2}A-\sec ^{2}A\cdot \csc ^{2}A$
$latex = \frac{1}{\cos ^{2}A}+\frac{1}{\sin ^{2}A}-\frac{1}{\sin ^{2}A\cdot \cos ^{2}A}&s=1$
$latex =\frac{\sin ^{2}A+ \cos ^{2}A-1}{\sin ^{2}A\cdot \cos ^{2}A}&s=1$
$latex = \frac{1-1}{\sin ^{2}A\cdot \cos ^{2}A} = 0&s=1$
Hence option [B] is correct answer.