Find $cot\space x_n$ if $sin(x_{n+1}-x_n)+2^{-(n+1)}sin\space x_nsin \space x_{n+1}=0$ for $n\ge1$ Also Given

Suppose $x_1=tan^{-1}2>x_2>x_3>......$ are positive real numbers

I cannot understand as to how I should approach but I have a hunch which is a bit hand-wavy way to prove it but tried to show that
$lim_{n \to \infty}x_n =\dfrac{\pi}{4}$ (For the proof see my edits in @hypernova's answer)
But can anybody prove it and then use it to find $cot \space x_n$ please?
 A: Note that
$$
\sin\left(x_{n+1}-x_n\right)=\sin x_{n+1}\cos x_n-\cos x_{n+1}\sin x_n.
$$
The iterative scheme thus reads
$$
\sin x_{n+1}\cos x_n-\cos x_{n+1}\sin x_n+2^{-n-1}\sin x_{n+1}\sin x_n=0.
$$
Thanks to the given condition that $x_j\in\left(0,\arctan 2\right]\subseteq\left(0,\pi\right)$, we have $\sin x_j\ne 0$. Hence the above scheme is equivalent to
$$
\cot x_n-\cot x_{n+1}+2^{-n-1}=0,
$$
or in a more friendly way,
$$
\cot x_{n+1}-\cot x_n=\frac{1}{2^{n+1}}.
$$
Therefore,
\begin{align}
\cot x_n&=\cot x_1+\sum_{j=2}^n\left(\cot x_j-\cot x_{j-1}\right)\\
&=\cot x_1+\sum_{j=2}^n\frac{1}{2^j}\\
&=\cot x_1+\frac{2^n-2}{2^{n+1}}\\
&=\cot\left(\arctan 2\right)+\frac{2^n-2}{2^{n+1}}\\
&=\frac{1}{2}+\frac{2^n-2}{2^{n+1}}\\
&=1-\frac{1}{2^n}.
\end{align}
For the proof of $lim_{n\to \infty}(x_n)=\dfrac{\pi}{4}$ which is infact a piece of cake now,
Using the above result we get,
$tan(x_n)=\dfrac{1}{1-\dfrac{1}{2^n}}=\dfrac{2^n}{2^n-1}$
$x_n= tan^{-1}\bigg(\dfrac{2^n}{2^n-1}\bigg)$
Now $lim_{n\to \infty}(x_n)=lim_{n\to \infty}\bigg(tan^{-1}\bigg(\dfrac{2^n}{2^n-1}\bigg)\bigg)$
$\implies \space lim_{n\to \infty}(x_n)=tan^{-1}(1)=\dfrac{\pi}{4}$ (Proved)
