I'm on pg 288 of Apostol's Calculus volume 1. A part of the theorem at the top says that 1/(1 + g(x)) = 1 - g(x) + o(g(x)) if g(x) -> 0 as x -> a. The first subsequent example is to show tan x = x +(1/3)x ^3 + o(x^3) as x -> 0. Using the theorem and the taylor approximation to cosine, we have 1/cos x = 1/(1 - (1/2)x^2 + o(x^3)) = 1 +(1/2)x^2 +o(x^2) as x ->0. I can't see how this follows from the theorem or the definition of o notation.