If `cos(theta - alpha) , costheta , cos(theta + alpha)` are in H.P. then `costheta.sec(alpha)/2 = `

Options

A. `sintheta=sqrt2cos((alpha)/(2))`
B. `costheta=sqrt2cos((alpha)/(2))`
C. `cos theta=sqrt2 sin((alpha)/(2))`
D. `sintheta=sqrt2sin((alpha)/(2))`

Answer

B

Explanation

It is given that `cos (theta-alpha), cos theta and cos (theta+alpha)` are in H.P. Therefore,
`(2)/(cos theta)=(1)/(cos(theta-alpha))+(1)/(cos(theta+alpha))`
`implies(2)/(cos theta)=(2costhetacosalpha)/(cos^(2)theta-sin^(2)alpha)`
`cos^(2)theta(1-cosalpha)=sin^(2)alpha`
`cos theta=sqrt2cos""(alpha)/(2)`