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

Maths 
Steve,
because the difference of an AP is constant,
sec(x)  sec(x+y) = sec(xy)  sec(x)
or, converting to cosines for ease of calculation,
cos(x+y) cos(xy)  cos(x) cos(xy) = cos(x) cos(x+y)  cos(x+y) cos(xy)
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)  (1cos^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)