Question

If A + B = 90°, then $\frac{tan AtanB+tanAtanB}{sinAsecB}-\frac{sin^{2}B }{cos^{2}A }$ is equal to
(a)$cot^{2}A$
(b)$cot^{2}B$
(c)$-tan^{2}A$
(d)$-cot^{2}A$

Answer 1

SOLUTION :  

The correct option is (b) : cot² B

Given : A + B = 90°  

B = 90° - A …………. (1)

A = 90° - B ………….(2)

The value of : [tan A tan B + tan A cot B / sin A sec B ]  - sin²B /cos²A

=[ tan A × tan (90° - A) + tan A cot(90° - A) / sinA sec(90° - A) ] - sin² (90° - A) / cos² A

[From eq 1]

= tan A cot A + tan A tan A  / sin A cosec A ]  - cos²A /cos²A

[ tan (90° - θ) = cot θ, cot (90° - θ) = tan θ, sec (90° - θ) = cosec θ]

=  1 + tan²A / 1 - 1

[ tan θ cot θ = 1 ,sin θ cosec θ = 1]  

= 1 + tan²A  - 1

=  tan²A  

= tan² (90° - B)      [ from eq 2]

= cot² B             [tan (90° - θ) = cot θ]

Hence, the value of : [tan A tan B + tan A cot B / sin A sec B ]  - sin²B /cos²A is cot² B .

HOPE THIS ANSWER WILL HELP YOU…

Answer 2

the correct answer is option (b)