Maclaurin's Series for $\sec(x)$ with help of Maclaurin's series for $\tan(x)$ 
Is there any way to derive Maclaurin's series for $\sec(x)$ with the help of Maclaurin's series for $\tan(x)$?

As we know, the Maclaurin's Series for $\tan(x)$ is:
$$\tan(x)=x+\frac{x^3}{3}+\frac{2x^5}{15}+....$$
 A: It is probably not the simplest way to do it but, using $$\frac{d}{dx}\tan(x)=\sec ^2(x)$$ you will need to compute $$\tan(x)=x+\frac{x^3}{3}+\frac{2 x^5}{15}+\frac{17 x^7}{315}+O\left(x^9\right)$$ $$\sec ^2(x)=1+x^2+\frac{2 x^4}{3}+\frac{17 x^6}{45}+O\left(x^8\right)$$ $$\sec(x)=\sqrt{1+x^2+\frac{2 x^4}{3}+\frac{17 x^6}{45}+O\left(x^8\right)}$$ Now, consider $$y=x^2+\frac{2 x^4}{3}+\frac{17 x^6}{45}+\cdots$$ $$\sqrt{1+y}=1+\frac{y}{2}-\frac{y^2}{8}+\frac{y^3}{16}-\frac{5 y^4}{128}+O\left(y^5\right)$$ and replace $y$ and expand.
Doing it with the terms given above, you would arrive to $$1+\frac{x^2}{2}+\frac{5 x^4}{24}+\frac{61 x^6}{720}-\frac{41
   x^8}{640}+\cdots$$ while the correct series should be $$1+\frac{x^2}{2}+\frac{5 x^4}{24}+\frac{61 x^6}{720}+\frac{277
   x^8}{8064}+\cdots$$
Edit
Another possible way to do it based on $$\int \tan(x)=\log(\sec(x))$$ So $$\log(\sec(x))=\frac{x^2}{2}+\frac{x^4}{12}+\frac{x^6}{45}+\frac{17 x^8}{2520}+O\left(x^{10}\right)$$ $$\sec(x)=\exp\left(\frac{x^2}{2}+\frac{x^4}{12}+\frac{x^6}{45}+\frac{17 x^8}{2520}+\cdots \right)$$ Now, consider $$y=\frac{x^2}{2}+\frac{x^4}{12}+\frac{x^6}{45}+\frac{17 x^8}{2520}+\cdots$$ and use $$\exp(y)=1+y+\frac{y^2}{2}+\frac{y^3}{6}+\frac{y^4}{24}+O\left(y^5\right)$$  Replace $y$ and expand.
Doing it with the terms given above, you would arrive to the correct series up to the $O\left(x^{9}\right)$.
Edit
Another way using the tangent half angle substitution. Let us start with $$\sec(x)=\frac{1+t^2}{1-t^2}=1+2 t^2+2 t^4+2 t^6+2 t^8+O\left(t^9\right)$$ using $$t=\tan(\frac x2)=\frac{x}{2}+\frac{x^3}{24}+\frac{x^5}{240}+\frac{17 x^7}{40320}+O\left(x^9\right)$$ Doing it with the terms given above, you would arrive to the correct series up to the $O\left(x^{9}\right)$.
