Find the first four nonzero terms of the Maclaurin expansion of f(x) = sec^2(x)

I have NO IDEA how to attempt this problem. I think you might have to do multiplication of power series but I don't even know how to do that.

Could you just find the derivatives and plug in zero?

You can use the derivatives and let x = 0, but this will become time-consuming in this problem.

Try this:
Note that cos^2(x) * sec^2(x) = 1.

Since sec^2(x) is even, we know that
sec^2(x) = A + Bx^2 + Cx^4 + Dx^6 + ...

As for cos^2(x), note that
cos^2(x) = (1/2)(1 + cos(2x))
.............= (1/2)[1 + (1 - (2x)^2/2! + (2x)^4/4! - (2x)^6/6! + ...)]
.............= 1 - x^2 + (1/3)x^4 - (2/45) x^6 + ...

Hence, cos^2(x) * sec^2(x) = 1
==> [1 - x^2 + (1/3)x^4 - (2/45) x^6 + ...] * [A + Bx^2 + Cx^4 + Dx^6 + ...] = 1.

Clearly A = 1 (looking at constant terms).

Expanding
[1 - x^2 + (1/3)x^4 - (2/45) x^6 + ...] * [1 + Bx^2 + Cx^4 + Dx^6 + ...] = 1 yields

1 + (-1 + B)x^2 + (C - B + 1/3)x^4 + (D - C + B/3 - 2/45)x^6 + ... = 1.

Equate like coefficients:
-1 + B = 0 ==> B = 1
C - B + 1/3 = C - 1 + 1/3 = 0 ==> C = 2/3.
D - C + B/3 - 2A/45 = D - 2/3 + 1/3 - 2/45 = 0 ==> D = 17/45.

Hence, sec^2(x) = 1 + x^2 + (2/3)x^4 + (17/45)x^6 + ...

Double check:
http://www.wolframalpha.com/input/?i=pow…

I hope this helps!