By itself, #sec(x)# is not easy to work with, since we haven't really spent a lot of time learning the values associated with it. How can we put it in the form of something simpler? Well, remember that:

#sec(x) = 1/cos(x)#

Hence, we'll have:

#sec^2(pi) = [1/cos(pi)]^2 = 1/cos^2(pi)#

You should know that #cos(pi) = -1# from the unit circle. Because we're squaring the #cos(x)#, the negative sign simply goes away, so we're left with:

#1/1 = color(red)(1)#

Check out the graph of #sec^2(x)# below. Notice how it has the coordinate #(pi, 1)#:

graph{y = (sec(x))^2 [-10, 10, -5, 5]}

Hope that helped :)