p. 513

68.

Plot[(2 - x^2)/x^2,  {x, -4, 4},  PlotRange -> {-2, 10}]

Plot[Cos[2  ArcSec[x]],  {x, -4, 4},  PlotRange -> {-2, 10}]

The graphs look the same. Mathematica can even simplify cos ((sec^(-1)) (x))to prove this:

TrigExpand[Cos[2  ArcSec[x]]]

To see this for yourself, draw a right triangle with hypothenuse equal to x and side adjacent to the angle α equal to 1. Then sec (α)  =  x and α  =  (sec^(-1)) (x). The side opposite α is (Sqrt[(x^2) -  1]). Using the double angle formula, we see that cos (2  (sec^(-1)) (x))  =  
        cos (2  α)  =  
          (cos^2) (α)  -  (sin^2) (α) = ((((1/x)))^2) -  (((((Sqrt[(x^2) - 1])/x)))^2) =  
        (1/(x^2)) -  (((x^2) -  1)/(x^2))  =  ((2  -  (x^2))/(x^2))