# MacLaurin Series of $\tan(x)$

I am trying to compute the MacLaurin series of $$\tan(x)$$. I know this one is $$\tan x=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}2^{2n}(2^{2n}-1)B_{2n}x^{2n-1}}{(2n)!}$$ And I know how to derive this formula. Indeed I simply express $$\tan$$ as a linear combination of $$\cot(x)$$ and $$\cot(2x)$$, for which we know the explicit formula $$\cot(x)=\sum_{n=0}^\infty \frac{(-1)^n2^{2n}B_{2n}x^{2n-1}}{(2n)!}$$ This formula is derived by writing $$\cot$$ in its exponential form and doing some algebra.

I know how to derive these formulas, but I do not understand what makes it the MacLaurin series for $$\tan(x)$$. Why couldn't they be any Taylor series centered somewhere else? And what even makes them a Taylor series, I only see it as a power series...

Thank you for you responses and help!

Because if we have a function $$f\colon(a-r,a+r)\longrightarrow\Bbb R$$ and a power series$$\sum_{n=0}^\infty a_n(x-a)^n\tag1$$centered at $$a$$ such that$$\bigl(\forall x\in(a-r,a+r)\bigr):f(x)=\sum_{n=0}^\infty a_n(x-a)^n,$$then automatically $$f$$ is a $$C^\infty$$ function and its Taylor series centered at $$a$$ is $$(1)$$.
For instance,$$|x|<1\implies\frac1{1-x}=\sum_{n=0}^\infty x^n$$and therefore the Taylor series of $$\frac1{1-x}$$ centered at $$0$$ is $$\sum_{n=0}^\infty x^n$$.