Suppose `theta and phi(ne 0)` are such that sec `(theta+phi), sec theta and sec (theta-phi)` are in A.P. If `cos theta=kcos((phi)/(2))` for some k, then k is equal to

Options

A. `+-1`
B. `+-2`
C. `+-sqrt2`
D. `+-(1)/(sqrt2)`

Answer

C

Explanation

It is given that `sec(theta+phi), sec theta and sec(theta-phi)` are in A.P.
`therefore2sec theta=sec (theta+phi)+sec(theta-phi)`
`implies(2)/(cos theta)=(cos (theta-phi)+sec(theta-phi))/(cos(theta+phi)cos(theta-phi))`
`implies(2)/(costheta)=(2costhetacosphi)/(cos^(2)theta-sin^(2)phi)`
`impliescos^(2)thetacos phi=cos^(2)theta-sin^(2)phi`
`impliescos^(2)theta(1-cosphi)=(1-cos^(2)phi)`
`impliescos^(2)theta=1+cos phi`
`impliescos^(2)theta=2cos^(2)""(phi)/(2)`
`impliescostheta=+-sqrt2cos""(phi)/(2)`
Hence, `k=+-sqrt2.`