If `sec alpha` is the average of `sec(alpha - 2beta) and sec(alpha + 2beta)` then the value of `(2 sin^2 beta - sin^2 alpha )` where `beta!= n pi` is

Answer

1

Explanation

`sec alpha = (sec ( alpha - 2 beta) + sec (alpha + 2 beta))/(2)`
`rArr = (2)/( cos alpha ) = ( cos ( alpha + 2 beta) +cos ( alpha - 2beta))/( cos ( alpha - 2 beta) cos ( alpha + 2beta )) `
` rArr cos 2alpha + cos 4 beta = cos alpha ( 2 cos alpha * cos 2 beta)`
`rArr 2 cos ^(2) alpha -1 + 2 cos ^(2) 2 beta - 1 = 2 cos ^(2) alpha cos 2beta `
` rArr cos^(2) alpha ( 1-cos 2 beta) + ( cos 2 beta + 1) ( cos 2 beta -1) =0`
`rArr (1-cos 2 beta) ( cos^(2) alpha - cos 2 beta -1) =0`
`rArr cos ^(2) alpha = cos 2 beta + 1" " ("as " beta ne n pi)`
` rArr cos ^(2) alpha = 2cos ^(2) beta`
1- sin ^(2) alpha = 2 ( 1-sin ^(2) beta)`
`rArr 2 sin ^(2) beta - sin ^(2) alpha =1 `