Question

cos(a-b), cosa, cos(a+b) are in H.P. then find value of cosa/cos(b/2)

Answer 1

as they are in H.P.

applying H.P. formula


    1/cos(a+b) – 1/cosa = 1/cosa – 1/cos(a-b)
 
   [cosa – cos(a+b)] / cos(a+b).cosa = [cos(a-b) – cosa] / cos(a-b).cosa
    cos(a-b)[cosa – cos(a+b)] = cos(a+b)[cos(a-b) – cosa]
    cos(a-b).cosa – [cos(a-b).cos(a+b)] = [cos(a-b).cos(a+b)] – cos(a+b).cosa
    cos(a-b).cosa + cos(a+b).cosa = 2.cos(a-b).cos(a+b)
    cosa [cos(a-b) + cos(a+b)] = 2(cos2a – sin2b)
    cosa.(2.cosa.cosb) = 2(cos2a – sin2b)
    cos2a.cosb – cos2a = -sin2b
    cos2a.(cosb – 1) = -(1 – cos2b)
    cos2a.(cosb – 1) = (cosb – 1)(cosb + 1)
    cos2a – 1 = cosb


Answer

cos(b/2)÷2cosa•cos^2(b/2)