A question on AP.

sec(x-y) secx sec(x+y) → AP
1cos(x-y) 1cos x 1cos(x+y) →AP

Therefore,

2cos x = 1cos(x-y) + 1cos(x+y)

2cos x = 2.cos x.cos ycos(x-y).cos(x+y)

1cos x = 2.cos x.cos ycos²x-sin²y

cos²x-sin²y = cos²x.cosy

cos²x (1 - cos y) = sin²y

cos²x (1 - cos y) = (1 - cos²y)

cos²x = 1 + cos y

cos²x - cos y = 1

Since maximum value of cos²x is (+1)
or cos y = ±1................................(1)
Therefore, cos y = 0 in order to fulfill the above equation.

cos y = 0

y= π2

y2 = π4

cos y2 = cosÏ€4 = 1√2

secy2 = √2.....................................(2)

Multiplying (1) and (2), we get
cos y . sec y2 = ±1 x √2 = ±√2 [ Option (d) ]

Thank you very much for the quick reply!