Question

If cos(x-y),cos(x),cos(x+y) are in H.P then the value of cox(x)sec(y/2) is ?

Answer 1

Because they are in Harmonic Progress (H.P.) we have

`1/cos(x+y)+1/cos(x-y)=2/cosx`

Hence

`cosx*[cos(x+y)+cos(x-y)]=2*cos(x+y)*cos(x-y)`

`cos^2x*cosy=cos^2x-sin^2y`

`cos^2x=cosy+1`

`cos² x = 2 cos² (y/2)`

`(cos²x) / (cos²(y/2)) = 2`

`cos² x. sec² (y/2) = 2`

Taking absolute values we have

`|cosx*sec(y/2)|=sqrt2`

or

`cosx*sec(y/2)=+-sqrt2`