Question

sec^2 theta -cot^2 (90-theta)/cosec^2 67-tan^2 23+(sin^2 40+sin^2 50

Answer 1

let theta be a
lets solve numerator first
sec^2a-cot^2(90-a)                          * { sec^2a -tan^2a=1}
= sec^2a-tan^a
=1
now denominator
=cosec^2 67-tan^2 23+(sin^2 40+sin^2 50)
=cosec^2 67-tan^2(90-67)+(sin^2 40+sin^2 90-40)  * { cosec^2 a-tan^2 a=1}
=cosec^2 67-cot^2 67+(sin^2 40+cos^2 40)              * {sin ^2  a+cos^2 a =1}
=1+1
=2
therefore answer is 1/2