`sin^(2)5+sin^(2)6^(@)+ . . . +sin^(2)84^(@)+sin^(2)85^(@)=?`

Options

A. `30(1)/(2)`
B. `40(1)/(2)`
C. `40`
D. `39(1)/(2)`

Answer

b

Explanation

`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)`