Problem source

Let 
\begin{alignat*}{2}
\tan x & =\ \ \, \quad{\displaystyle \sum_{n=0}^{\infty}a_{n}x^{n}} &  & \text{ for small }x,\\
x\cot x & =1+\sum_{n=1}^{\infty}b_{n}x^{n}\quad &  & \text{ for small }x\text{ and not zero}.
\end{alignat*}
Using the relation 
\[
\cot x-\tan x=2\cot2x,\tag{*}
\]
 or otherwise, prove that $a_{n-1}=(1-2^{n})b_{n}$, for $n\geqslant1$. 
Let 
\[
x\mathrm{cosec}x=1+{\displaystyle \sum_{n=1}^{\infty}c_{n}x^{n}\quad\text{ for small }x\neq0. \qquad \qquad \, }
\]
Using a relation similar to $(*)$ involving $2\mathrm{cosec}2x$, or otherwise, prove that 
\[
c_{n}=\frac{2^{n-1}-1}{2^{n}-1}\frac{1}{2^{n-1}}a_{n-1}\qquad(n\geqslant1).
\]

Solution source

\begin{align*}
&& \cot x - \tan x &= 2 \cot 2x \\
\Rightarrow && x\cot x - x\tan x &= 2x\cot 2x \\
\Rightarrow && 1 + \sum_{n=1}^{\infty} b_n x^n - \sum_{n=0}^{\infty}a_n x^{n+1} &= 1 + \sum_{n=1}^{\infty} b_n (2x)^n \\
\Rightarrow  && \sum_{n=1}^{\infty}(1-2^n)b_nx^n &= \sum_{n=1}^{\infty} a_{n-1}x^n \\
\Rightarrow && a_{n-1} &= (1-2^n)b_n \quad \text{if }n \geq 1
\end{align*}


\begin{align*}
\cot x + \tan x &= 2 \cosec 2x 
\end{align*}

So
\begin{align*}
&& \cot x + \tan x &= 2 \cosec 2x  \\
\Rightarrow && x \cot x + x\tan x &= 2x \cosec 2x  \\
\Rightarrow &&  1 + \sum_{n=1}^{\infty} b_n x^n + \sum_{n=0}^{\infty} a_n x^{n+1} &= 1+\sum_{n=1}^\infty c_n (2x)^n \\
\Rightarrow && \sum_{n=1}^{\infty} \frac{1}{1-2^n}a_{n-1} +\sum_{n=1}^{\infty}a_{n-1}x^n &= \sum_{n=1}^{\infty} 2^nc_n x^n \\
\Rightarrow && c_n &= \frac{1}{2^n} \left ( 1 + \frac{1}{1-2^n} \right)a_{n-1} \\
&&&= \frac1{2^n} \frac{2^n-2}{2^n-1} a_{n-1}\\
&&&= \frac1{2^{n-1}}\frac{2^{n-1}-1}{2^n-1} a_{n-1}
\end{align*}