Let's consider what the triangle we're doing trigonometric functions might look like: 
If we consider that #tan(A)=b/a#, and we know that #tan(A)=t=t/1#, we can say that #a=1# and #b=t#. If we solve for the remaining side using the Pythagorean Theorem, we get #c=sqrt(t^2+1)#.
Now we know all the sides in terms of t, we can just compute the secant. Since the #sec(A)=c/a#, in this context we get that #sec(A)=sqrt(t^2+1)/1=sqrt(t^2+1)#
We have gotten to the answer already, but I want to elaborate on a step we took in the middle. This step confused me a lot when I first did something like this, and it is where we just assume that the fraction of #b/a=t/1#. I always wondered how one could know that it was #t/1#, and not some other of the infinitely many other possible fractions. The answer is that #t/1# probably wasn't the original fraction used to compute the tangent, but it doesn't matter because the triangle with the sides #t# and #1# will be similar to the triangles with sides of all of the other possible fractions, and therefor the trigonometric calculations will have the same results, so it doesn't matter that we picked a side to seemingly arbitrarily be #1#
