Show that the solutions of the equation $(1+x)^{2n}+(1-x)^{2n}=0$ are $x=\pm i ~\tan{\frac{(2r-1)\pi}{4n}}, ~~r=1,2,\cdots n.$ 
Show that the solutions of the equation $(1+x)^{2n}+(1-x)^{2n}=0$ are $$x=\pm i~ \tan{\frac{(2r-1)\pi}{4n}}, ~~r=1,2,\cdots n.$$

Attempt
Clearly, $$\frac{1+x}{1-x}=(-1)^{1/2n}=(\cos{(2k+1)\pi}+i \sin{(2k+1)\pi})^{1/2n}=\cos{\frac{(2k+1)\pi}{2n}}+i \sin{\frac{(2k+1)\pi}{2n}}$$
(by de Moivre's formula)
By componendo dividendo and some simplification,
I am getting $$x=i\tan{\frac{2k+1}{4n}\pi}, ~~k=1,2,3,\cdots 2n$$
How to get the desired result? I am getting 2n solutions and I understand that the solutions that I have obtained is equal to that of the desired. But how to to get the desired from the solution that I have obtained? Please provide me the mathematical steps.
Is there any other method of solving to get the answer directly. 
 A: From
$$
\frac{1+x}{1-x}=\left[\pm(\sqrt{-1})\right]^{1/n}=(\pm i)^{1/n}=\mathrm e^{\frac{i}{n}(\pm\pi/2+k\pi)}
$$
using the fact that
$$
\begin{align*}
i&=\mathrm e^{i\pi/2}=\underbrace{\cos(\pi/2)}_{0}+i\sin(\pi/2)\\
-i&=-\mathrm e^{i\pi/2}=\underbrace{-\cos(\pi/2)}_{0}-i\sin(\pi/2)=\mathrm e^{-i\pi/2}\\
\Longrightarrow \pm i&=\mathrm e^{\pm i\pi/2}
\end{align*}
$$
and observing that after $k\pi$ we have a periodical change from $i$ to $-i$, one has
$
\pm i=\mathrm e^{i(\pm \pi/2+k\pi)}
$.
Now put $a=\pm \pi/2+k\pi$. We have for $+a$
$$
x=\frac{-1+\mathrm e^{ia}}{1+\mathrm e^{ia}}
=\frac{\mathrm e^{ia/2}}{\mathrm e^{ia/2}}\cdot\frac{\mathrm e^{ia/2}-\mathrm e^{-ia/2}}{\mathrm e^{ia/2}+\mathrm e^{-ia/2}}=\frac{i\sin\left({\frac{a}{2}}\right)}{\cos\left({\frac{a}{2}}\right)}=i\tan\left(\frac{a}{2}\right)
$$
and for $-a$, $x=-i\tan\left(\frac{a}{2}\right)$
that is
$$
x=\pm i\tan\left(\frac{2k+1}{4n}\pi\right)\quad \text{for }k=0,1,2,\ldots
$$
or

$$
x=\pm i\tan\left(\frac{2r-1}{4n}\pi\right)\quad \text{for }r=1,2,\ldots
$$

A: For any $\alpha$, $\tan\alpha=\tan(\alpha-\pi)$.
For $n\leq k<2n$:
$$\begin{align}i\tan\frac{2k+1}{4n}\pi &= i\tan\frac{2(k-2n)+1}{4n}{\pi}
\\
&=-i\tan\frac{2(2n-k)-1}{4n}\pi
\end{align}$$
So $k=n,n+1,\dots,2n-1$ corresponds to $r=2n-k$ and the minus sign.
$k=2n$ corresponds to $r=1$ with the positive sign.
$k=1,2,\dots,n-1$ corresponds to $r=k+1$ with the positive sign.
