Question

If A + B = 90°, then \[\frac{\tan A \tan B + \tan A \cot B}{\sin A \sec B} - \frac{\sin^2 B}{\cos^2 A}\] 

पर्याय

• cot² A

• cot² B

• −tan² A

• −cot² A

Answer 1

We have: 

`A+B=90°` 

`⇒ B=90°-A` 

We have to find the value of the following expression

`(tan A tan B+tan A cot B)/(sin A sec B)-sin ^2 B/cos^2 A` 

so 

`(tan A tan B+tan A cot B )/(sin A sec B)- sin^2 B/cos^2 A` 

`(tan A tan (90°-A)+ tan A cot (90°-A))/(sin A sec (90°-A))-sin^2(90°-A)/cos^2 A` 

=` (tan A cot A + tan A tan A)/(sin A cosec A)-cos^2A/cos^2 A` 

= `1+tan^2 A-1`

=` tan ^2 A` 

=` tan ^2(90°-B)` 

=` cot^2 B`