Correct Answer - b
`sin^(2)5^(@)+sin^(2)85^(@))+(sin^(2)6^(@)+ sin^(2)84^(@))+`............upto 40 pairs+middle term =40`+sin^(2)45^(@)`
`=40+(1)/(2)=40(1)/(2)`
Alternate In case, when series is in the form of `sin^(2)theta` or `cos^(2)theta`, then sum of series will always be half of no. of terms.
`=(81)/(2=40(1)/(2)`