Using the Maclaurin Series of $\ln(\sec x)$ find the first two non-zero terms of $\tan x$

Question

This is easy using the theorem but how can I use $\ln(\sec x)$, which is $\frac{x^2}2+\frac{x^4}{12}+\dots$ to find the series of $\tan x$?

I thought about finding the series of $\ln(\sec (pi/2-x))$ then subtracting the two series to find $\ln(\tan x)$ and then e to the power of that series to get tanx. But there has to be an easier way than this.

Answer 1

Hint: $\frac{\mathrm{d}}{\mathrm{d}x}\ln\sec x=\tan x$