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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