We know, the function $arctan(x)$ is defined as the real number $y$ such that $tan(y)=x$ and $-\frac{\pi}{2}\leq y\leq\frac{\pi}{2}$
From this definition we can derive the well-known result:$$\frac{d}{dx}arctan(x)=\frac{1}{1+x^2}$$
Let another function $arctan_2(x)$ be defined as the real number $y$ such that $tan(y)=x$ and $\frac{3\pi}{2}\leq y\leq\frac{5\pi}{2}$
[We can observe that the function is single-valued over $[\frac{3\pi}{2},\frac{5\pi}{2}]$ and so the one-to-one property remains there]
Again, by definition, $$\frac{d}{dx}arctan_2(x)=\frac{1}{1+x^2}$$
So, we'll obtain the same Taylor expansion$$x-\frac{x^3}{3}+\frac{x^5}{5}-...$$
for both the functions $arctan(x)$ and $arctan_2(x)$, when $-1\leq x\leq 1$
Thus, for any real $a$, given $-1\leq a\leq 1$,$$arctan(a)=a-\frac{a^3}{3}+\frac{a^5}{5}-...=arctan_2(a)$$
But $-\frac{\pi}{2}\leq arctan(a)\leq\frac{\pi}{2}$ whereas $\frac{3\pi}{2}\leq arctan_2(a)\leq\frac{5\pi}{2}$
i.e. $arctan(a)$ can never be equal to $arctan_2(a)$ !!!!!
At which point am I going wrong?
You have to remember the pesky "$+C$" that shows up when you take indefinite integrals.
When you integrate the series for $\frac{1}{1+x^2}$ to get a series for $\arctan(x)$ or for $\arctan_2(x)$, you don't get
$$x-\frac{x^3}3 + \frac{x^5}5 - \dots$$
but instead
$$C + x - \frac{x^3}3 + \frac{x^5}5 - \dots$$
This $C$ value is determined as $\arctan(0)$ or $\arctan_2(0)$, which are different, and so result in different series.