Let $f(x)=\arcsin(1-x)$ for $x\in[0,2]$.
Since the derivative of $f(x)=O\left( x^{-1/2}\right)$ for $x\sim 0$, we let $t=x^{1/2}$ and $g(t)=\arcsin(1-t^2)$.
We will now develop the first few terms of the Taylor series for $g(t)$ around $t=0$.
We have for the first derivative $g^{(1)}(t)$
$$\begin{align}
g^{(1)}(t)&=-\frac{2t}{\sqrt{1-(1-t^2)^2}}\\\\
&=-\frac{2}{\sqrt{2-t^2}}\tag 1
\end{align}$$
Differentiating the right-hand side of $(1)$, we obtain the second derivative, $g^{(2)}(t)$
$$\begin{align}
g^{(2)}(t)&=-\frac{2t}{(2-t^2)^{3/2}}\tag 2
\end{align}$$
Continuing, we have for $g^{(3)}(t)$
$$\begin{align}
g^{(3)}(t)&=-\frac{4(t^2+1)}{(2-t^2)^{5/2}}\tag 3
\end{align}$$
And finally, we have for $g^{(4)}(t)$
$$\begin{align}
g^{(4)}(t)&=-\frac{12t(t^2+3)}{(2-t^2)^{7/2}}\tag 4
\end{align}$$
We evaluate $(1)-(4)$ at $t=0$ and form the expansion
$$\bbox[5px,border:2px solid #C0A000]{\arcsin(1-x)=\frac{\pi}{2}-\sqrt{2}x^{1/2}-\frac{\sqrt{2}}{12}x^{3/2}+O\left(x^{5/2}\right)}$$
The exponential generating function for the Bernoulli Numbers is
$$
\sum_{n=0}^\infty\frac{B_nx^n}{n!}=\frac{x}{e^x-1}\tag{1}
$$
The even part of $(1)$ is
$$
\sum_{n=0}^\infty\frac{B_{2n}x^{2n}}{(2n)!}=\frac x2\coth\left(\frac x2\right)\tag{2}
$$
Since $\cot(x)=i\coth(ix)$, by substituting $x\mapsto ix$, we get
$$
\begin{align}
\sum_{n=0}^\infty\frac{(-1)^nB_{2n}x^{2n}}{(2n)!}
&=\frac{ix}2\coth\left(\frac{ix}2\right)\\
&=\frac x2\cot\left(\frac x2\right)\tag{3}
\end{align}
$$
Therefore,
$$
\cot(x)=\sum_{n=0}^\infty\frac{(-1)^nB_{2n}2^{2n}x^{2n-1}}{(2n)!}\tag{4}
$$
Since $\tan(x)=\cot(x)-2\cot(2x)$, we get
$$
\begin{align}
\tan(x)
&=\sum_{n=0}^\infty\frac{(-1)^nB_{2n}2^{2n}\left(x^{2n-1}-2\cdot2^{2n-1}x^{2n-1}\right)}{(2n)!}\\
&=\sum_{n=1}^\infty\frac{(-1)^{n-1}B_{2n}\left(2^{2n}-1\right)2^{2n}x^{2n-1}}{(2n)!}\tag{5}
\end{align}
$$
Best Answer
You may calculate derivative using $\tan=\sin/\cos$, as comment say. But you can also try the following strategy.
Let $\tan x=\sum_n a_nx^n$
you know that the derivative of $\tan x$ equals $1+(\tan x)^2$
So you know
$$\sum_n na_nx^{n-1}=1+(\sum_i a_ix^i)(\sum_i a_i x^i)=1+\sum_n(\sum_{k+m=n-1}a_ka_m)x^n$$
You also knonw that $\tan(-x)=-\tan x$, whence $a_n=0$ for $n$ even.
Using this you can arrange an inductive process to calculate the taylor expansion. (Namely, $a_{2k}=0, a_1=1, a_3=1/3, a_5=2/15...$ and so on)
See also http://en.wikipedia.org/wiki/Taylor_series and http://en.wikipedia.org/wiki/Bernoulli_numbers for the explicit result