Question

Evaluate the following. `(sec^2(90-theta)-cot^2 theta)/(2(sin^(2)25^@+sin^(2) 65^@))-(2cos^(2)60 tan^(2)28^@ tan^(2)62)/(3(sec^2 43^@-cot^(2)47^@))`

Answer 1

`(sec^2(90-theta)-cot^2theta)/(2(sin^2 25^@+sin^2 65^@)) - (2cos^2 60^@tan^2 28^@tan^2 62^@)/(3(sec^2 43^@ - cot^2 47^@))`
`=(cosec^2theta-cot^2theta)/(2(sin^2 25^@+sin^2 (90-25)^@)) - (2cos^2 60^@tan^2 (90-62)^@tan^2 62^@)/(3(sec^2 43^@ - cot^2 (90-43)^@))`
`=(1)/(2(sin^2 25^@+cos^2 25^@)) - (2(1/2)^2 cot^2 62^@tan^2 62^@)/(3(sec^2 43^@ - tan^2 43^@))`
`=1/2-(1/2)/(3)`
`=1/2-1/6`
`=2/6 = 1/3`
`:. (sec^2(90-theta)-cot^2theta)/(2(sin^2 25^@+sin^2 65^@)) - (2cos^2 60^@tan^2 28^@tan^2 62^@)/(3(sec^2 43^@ - cot^2 47^@)) = 1/3`