Let cosx = y

=> x = cos^-1 y ....1

Again

 Let cosx = y

=> 1/cosx = 1/y

=> secx = 1/y          (cosx = 1/secx => secx = 1/cosx)

=> x =  sec^-1 (1/y) .........2

From equation 1 and 2, we get

  cos^-1 y = sec^-1 (1/y)

So cos inverse is the reciprocal of the secant inverse.