If `cos (x-y),, cosx and cos(x+y)` are in H.P., are in H.P., then `cosx*sec(y/2)`=

Options

A. `-sqrt(3)`
B. `-sqrt(2)`
C. `sqrt(2)`
D. `sqrt(3)`

Answer

B, C

Explanation

We have,
`" " (2)/(cosx ) = (1)/(cos(x-y)) + (1)/(cos(x+y)) = ( 2cosx*cosy)/( cos^(2)x - sin^(2)y)`
`rArr cos^(2) x - sin^(2)y = cos^(2) x * cos y`
`rArr cos^(2)x ( 1-cosy) = sin^(2)y`
`rArr cos^(2) x* 2 sin^(2)""(y)/(2) = 4 sin^(2) ""(y)/(2)cos^(2)""(y)/(2)`
`rArr cos ^(2)x= 2 cos^(2)""(y)/(2)`
`rArr cos^(2)x sec^(2) ""(y)/(2)= 2`
`rArr cosx * sec""(y)/(2) = pm sqrt2`