If sec(x+y),sec(x),sec(x-y) are in A.P. then prove that cosx= +-√2 cos y/2 where cosx and cosy are not equals to 1,

Answer 1

because the difference of an AP is constant,

sec(x) - sec(x+y) = sec(x-y) - sec(x)

or, converting to cosines for ease of calculation,

cos(x+y) cos(x-y) - cos(x) cos(x-y) = cos(x) cos(x+y) - cos(x+y) cos(x-y)

2(cosx cosy - sinx siny)(cosx cosy + sinx siny) = cosx(cosx cosy + sinx siny) + cosx(cosx cosy - sinx siny)

cos^2(x)cos^2(y) - sin^2(x)sin^2(y) = cos^2(x)cosy

cos^2(x)cos^2(y) - (1-cos^2(x)-cos^2(y)+cos^2(x)cos^2(y)) = cos^2(x)cos(y)

cos^2(x)(cos(y)-1) = cos^2(y)-1
cos^2(x) = cos(y)+1
cos^2(x) = 2cos^2(y/2)
cos(x) = ±√2 cos(y/2)

Answer 2

Thanks yaar