Find the first three terms of the Maclaurin Series 
Determine using multiplication/division of power series (and not via WolframAlpha!) the first three terms in the Maclaurin series for $y=\sec x$.

I tried to do it for $\tan(x)$ but then got kind of stuck. For our homework we have to do it for the $\sec(x)$. It is kind of tricky. Help would be awesome! 
Thanks!
Taylor series for $\tan(x)$:
\begin{align*}
\tan (x)
&=\frac{\sin(x)}{\cos(x)}\\
&=\frac{x-\frac {x^3}6+\frac{x^5}{120}-\cdots}{1-\frac{x^2}2+\frac{x^4}{24}-\cdots}\\
&=x+\frac{x^3}3+\frac{2x^5}{15}+\cdots
\end{align*}
 A: Let's write
$$1 = \cos{x} \sec{x} = \left(1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots\right)(a_0 + a_1 x + a_2x^2 + \dots)$$
Expand the right hand side to find that
$$1 = 1 (a_0) + x (a_1) + x^2 \left(a_2 - a_0 \frac{1}{2!}\right) + x^3 \left(a_3 - a_1 \frac{1}{2!}\right) + x^4 \left(a_4 - a_2 \frac{1}{2!} + a_0 \frac{1}{4!}\right)+\dots$$
Equating coefficients gives (note that the left side is $1 + 0x + 0x^2 + 0x^3 + \dots$)
\begin{align*}
1 &= a_0 \\
0 &= a_1 \\
0 &= a_2 - \frac{a_0}{2} \\
0 &= a_3 - \frac{a_1}{2} \\
0 &= a_4 - \frac{a_2}{2} + \frac{a_0}{24}
\end{align*}
Solving this gives $a_0 = 1$, $a_1 = 0 = a_3$, $a_2 = \frac{1}{2}$ and $a_4 = \frac{5}{24}$, or
$$\boxed{\sec{x} \approx 1 + \frac{1}{2} x^2 + \frac{5}{24} x^4}$$
A: $\sec(x)=\frac{1}{\cos x}$. The three first terms are $1,x^2$ and $x^4$. Then we  write:  $$\cos x=1-\frac{x^2}2 + \frac{x^4}{24} + o(x^4)$$
Putting $$u=-\frac{x^2}2 + \frac{x^4}{24}=-\frac{x^2}{2}\left(1-\frac{x^2}{12}\right)$$ we have:
$$\sec(x)=(1+u)^{-1} =\operatorname{Tronc}_4 (1 -u +u^2-u^3+u^4)=\operatorname{Tronc}_4 (1 -u +u^2)$$
Since: $\operatorname{Tronc}_4(u)=u$  and $\operatorname{Tronc}_4(u^2)=\frac{x^4}{4}$ you can finish:
$$\sec x = 1+\frac{x^2}{2}-\frac{x^4}{24} + \frac{x^4}4 + o(x^4)= 1+\frac{x^2}{2} + \frac{5x^4}4 + o(x^4) $$
