Please, help me to find where is a mistake in the solutions of the equation. I have this equation and I will be very thankful to anyone who can provide me any help with the one discrepancy in my solution and the solution from the self-learning website:
$$
\frac{1+\tan(x) + \tan^2(x) + ... + \tan^n(x) + ...}{1-\tan(x) + \tan^2(x) - ... +(-1)^n\tan^n(x) ...}= 1+\sin(2x)
$$

My solution
I solved it in not too graceful way. First I found the roots for this equation:
$$
\require{cancel}
\begin{align}
&\boxed{\frac{1+\tan(x)}{1-\tan(x)}= 1+\sin(2x)} \\ \\
&\frac{(1+\tan(x))\cancel{(1-\tan(x))}}{\cancel{1-\tan(x)}}= (1+\sin(2x))(1-\tan(x))\\ \\
&\cancel{1}+\tan(x)=\cancel{1} - \tan(x) + \sin(2x) - \sin(2x)\tan(x) \\
&\frac{\sin(2x)}{\tan(x)} - \frac{\sin(2x)\cancel{\tan(x)}}{\cancel{\tan(x)}} -\frac{2\cancel{\tan(x)}}{\cancel{\tan(x)}} = 0\\ \\
&\bbox[Beige]{\boxed{\sin(2\alpha)=2\sin(\alpha)\cos(\alpha)}} \\ \\
&\frac{\cancel{2}\cancel{\sin(x)}\cos(x)\cos(x)}{\cancel{\sin(x)}} - \cancel{2}\sin(x)\cos(x) - \cancel{2} = 0 \\ \\
&\cos^2(x) - \sin(x)\cos(x) -1 = 0\\ \\
&\bbox[Beige]{\boxed{\sin^2(\alpha) + \cos^2(\alpha) = 1}} \\ \\
&\cancel{\cos^2(x)} - \sin(x)\cos(x) - \sin^2(x) - \cancel{\cos^2(x)} = 0\\ \\
&\sin^2(x) + \sin(x)\cos(x) = 0 \\ \\
&\sin(x)(\sin(x) + \cos(x)) = 0 \\ \\
&\sin(x) = 0 \implies \boxed{x_1=\pi k ,\space k \in \mathbb{Z}} \\ \\
&\sin(x) + \cos(x) = 0 \implies \tan(x) = -1 \implies \boxed{x_2=\frac{\pi}{4}\left(4m - 1\right) ,\space m \in \mathbb{Z}}
\end{align}
$$
Also I checked that there were no parasite and missing roots, cause I've divided the equation by $\tan(x)$ and multiplied it by $1-\tan(x)$. There were no such ones cause $\tan(x)$ cannot assume $1$, when $\sin(x)=0$ or $\tan(x) = -1$, and the missing roots with $\tan(x) = 0$ are a subset of $x_1$.
Then I checked which of these roots fitted the next equation:
$$
\frac{1+\tan(x) + \tan^2(x)}{1-\tan(x) + \tan^2(x)}= 1+\sin(2x)
$$
And it turned out that only $x_1$ did. Then my logic was: if there were another roots for the next steps of this sequence, e.g. $\frac{1+\tan(x) + \tan^2(x) + \tan^3(x)}{1-\tan(x) + \tan^2(x) - \tan^3(x)}= 1+\sin(2x)$ they would not fit the first step while $x_1$ would fit all steps. So the only solution is $x_1 = \pi k ,\space k \in \mathbb{Z}$

Possibly the wrong solution with my correction. You can see the original one at the www.bymath.com

It is obvious that the fraction numerator and denominator are geometrical sequences (progressions) with the common ratios $\tan(x)$ and $-\tan(x)$ correspondingly. Note, that here $|\tan(x)| < 1$, otherwise the left-hand side expression is meaningless. Therefore it is possible to transform the fraction numerator and denominator by the sum formula for the unboundedly decreasing geometric sequence (progression) $\bbox[Beige]{\boxed{S = \frac{b_1}{1-q}}}$, where $b$ is the first member of a sequence and $q$ in a ratio.

$$
\begin{align}
&\frac{1}{1 - \tan(x)} : \frac{1}{1 - (-\tan(x))} = 1 + \sin(2x) \\
&\frac{1}{1 - \tan(x)} * (1 + \tan(x)) = 1 + \sin(2x) \\
&\bbox[Beige]{\boxed{\sin(\alpha) = \frac{2\tan(\frac{\alpha}{2})}{1+\tan^2(\frac{\alpha}{2})}}} \\
&\frac{1 + \tan(x)}{1 - \tan(x)} - 1 - \frac{2\tan(x)}{1+\tan^2(x)} = 0\\ \\
&\frac{(1 + \tan(x))(1+\tan^2(x)) - (1 - \tan(x))(1+\tan^2(x)) - 2\tan(x)(1 - \tan(x))}{(1 - \tan(x))(1+\tan^2(x))} = 0\\ \\
&\frac{1 + \tan^2 x + \tan x + \tan^3 x  -(1 + \tan^2 x - \tan x - \tan^3 x ) -2\tan x + 2\tan^2 x}{(1 - \tan x)(1+\tan^2 x)} =0 \\
\end{align}
$$
So here is the point where this second solution is wrong. The right is this:
$$
\begin{align}
&\frac{\cancel{1} + \cancel{\tan^2 x} + \cancel{\tan x} + \tan^3 x -\cancel{1} - \cancel{\tan^2 x} + \cancel{\tan x} + \tan^3 x -\cancel{2\tan x} + 2\tan^2 x}{(1 - \tan x)(1+\tan^2 x)} =0 \\
&\frac{2\tan^3 x + 2\tan^2 x}{(1 - \tan x)(1+\tan^2 x)} =0 \implies \tan x = 0, \ \tan x = -1 \\
\end{align}
$$
Whereas the author of the site got this: $\frac{4\tan^3 x}{(1 - \tan x)(1+\tan^2 x)} =0 \implies \tan x = 0$ and lost one root $x_2=\frac{\pi}{4}\left(4m - 1\right) ,\space m \in \mathbb{Z}$

ACTUALLY QUESTION
So when I looked to the second solution I was pretty confused with this:
in that solution it is said that unless $|tan(x)| < 1$ the left-hand side of the equation in meaningless. Why, can somebody explain, please? I suppose that this is not true cause:


*

*while the transformation with the sum formula was correct, the solution itself was wrong

*the right solution gives us two roots one of which does not fit all possible steps of the sequences, hence the numerator and denominator are not a  unboundedly decreasing geometric sequence (progression).


Is my conclusion correct?
My sincerest appreciation.

If $|\tan x|\ge 1$ both numerator and denominator of the left-hand side are not finite, because the geometric series 
$$N=1+\tan(x) + \tan^2(x) + ... + \tan^n(x) + ...$$
and
$$D=1-\tan(x) + \tan^2(x) - ... +(-1)^n\tan^n(x)+...$$
don't converge. If $|\tan x|< 1$, we have
$$N=\lim_{n\rightarrow \infty }\frac{\tan ^{n}x-1}{\tan x-1}=\frac{0-1}{\tan x-1}=\frac{1}{1-\tan x}$$
and
$$D=\lim_{n\rightarrow \infty }\frac{(-\tan x)^{n}-1}{(-\tan x)-1}=\frac{0-1}{-\tan x-1}=\frac{1}{1+\tan x};$$
and the given equation is equivalent to
$$\begin{eqnarray*}
\frac{1+\tan x}{1-\tan x} &=&1+\frac{2\tan x}{1+\tan ^{2}x} \\
\frac{1+\tan x}{1-\tan x}&=&\frac{(1+\tan ^{2}x)^{2}}{1+\tan ^{2}x},\qquad |\tan x|< 1,\tag{1}
\end{eqnarray*}$$
because 
$$\sin 2x=\frac{2\tan x}{1+\tan ^{2}x}.$$
To solve equation $(1)$ we can observe that it is equivalent to
$$\frac{\left( 1+\tan x\right) ( 1+\tan ^{2}x) }{(1-\tan
x) \left( 1+\tan ^{2}x\right) }=\frac{(1+\tan x)^{2}\left( 1-\tan
x\right) }{\left( 1-\tan x\right) \left( 1+\tan ^{2}x\right) },\qquad |\tan x|< 1,$$
and to
$$\begin{eqnarray*}
\left( 1+\tan x\right) ( 1+\tan ^{2}x)  &=&(1+\tan x)^{2}\left(
1-\tan x\right) \\
&\Leftrightarrow &1+\tan ^{2}x=(1+\tan x)\left( 1-\tan x\right)  \\
&\Leftrightarrow &1+\tan ^{2}x=1-\tan ^{2}x \\
&\Leftrightarrow &\tan ^{2}x=-\tan ^{2}x \\
&\Leftrightarrow &2\tan ^{2}x=0 \\
&\Leftrightarrow &\tan x=0,
\end{eqnarray*}$$
whose single solution is $x_1=k\pi,k\in\mathbb{Z}$.
