Maclaurin polynomial of tan(x) The method used to find the Maclaurin polynomial of sin(x), cos(x), and $e^x$ requires finding several derivatives of the function. However, you can only take a couple derivatives of tan(x) before it becomes unbearable to calculate.
Is there a relatively easy way to find the Maclaurin polynomial of tan(x)? 
I considered using tan(x)=sin(x)/cos(x) somehow, but I couldn't figure out how.

Long division of series.
$$ \matrix{ & x + \frac{x^3}{3} + \frac{2 x^5}{15} + \dots
\cr 1 - \frac{x^2}{2} + \frac{x^4}{24} + \ldots & ) \overline{x - \frac{x^3}{6} + \frac{x^5}{120} + \dots}\cr
& x - \frac{x^3}{2} + \frac{x^5}{24} + \dots\cr & --------\cr
& 
\frac{x^3}{3} - \frac{x^5}{30} + \dots\cr
& \frac{x^3}{3} - \frac{x^5}{6} + \dots\cr
& ------\cr
&\frac{2 x^5}{15} + \dots \cr
&\frac{2 x^5}{15} + \dots \cr
& ----}$$
