If `cos(x-y),cosx and "cos"(x+y)` are in H.P., then `cosxsec(y/2) is`

Options

A. `pmsqrt(2)`
B. `(1)/(sqrt(2))`
C. `-(1)/(sqrt(2))`
D. None of these

Answer

A

Explanation

`:.cos(x-y),cosx,cos(x+y)` are in HP.
`:.cosx=(2cos(x-y)cos(x+y))/(cos(x-y)+cos(x+y))`
`implies cosx=(2cos^(2)x-sin^(2)y)/(2cosx cosy)`
`implies cos^(2)x cosy =cos^(2)x-sin^(2)y`
`implies cos^(2)x(1-cosy)=sin^(2)y`
`=(1+cos y)(1-cosy )`
`implies cos^(2)x=(1+cosy)" "[:.1-cosyne0]`
`implies cos^(2)x= 2 cos^(2)""(y)/(2)`
`implies cos^(2)xsec^(2)((y)/(2))=2`
`:. cos x sec ((y)/(2))= pm sqrt(2)`