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

1 Answer

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#