Problem source

\begin{questionparts}
\item
Let 
\[
\tan x = \sum\limits_{n=0}^\infty a_n x^n
\text{ and }
\cot x = \dfrac 1 x +\sum\limits_{n=0}^\infty b_nx^n
\]
for $0< x  < \frac12\pi\,$. Explain why $a_n=0$ for even $n$.
Prove  the identity
\[
\cot x - \tan x \equiv 2 \cot 2x\,
\]
and show that \[a_{n} = (1-2^{n+1})b_n\,.\]
\item
Let 
$ \displaystyle {\rm cosec}\, x 
= \frac1x +\sum\limits _{n=0}^\infty c_n x^n\,$ for 
$0< x < \frac12\pi\,$.
By considering $\cot x + \tan x$, or otherwise, show that
\[
c_n = (2^{-n} -1)b_n
\,.
\]
\item Show that 
\[
\left(1+x{ \sum\limits_{n=0}^\infty} b_n x^n \right)^2 +x^2
= \left(1+x{ \sum\limits_{n=0} ^\infty} c_n x^n \right)^2\,.
\]
Deduce from this and the previous results that $a_1=1$, and find $a_3$. 
\end{questionparts}

Solution source

\begin{questionparts}
\item Since $\tan (-x) = -\tan x$, $\tan$ is an odd function, and in particular all it's even coefficients are zero. 

\begin{align*}
&& 2 \cot 2x &\equiv \frac{2 cos 2x}{\sin 2 x} \\
&&&\equiv \frac{2(\cos^2 x- \sin^2 x)}{2 \sin x \cos x} \\
&&&\equiv \frac{\cos x}{\sin x} - \frac{\sin x}{ \cos x} \\
&&&\equiv \cot x - \tan x
\end{align*}

Therefore

\begin{align*}
&& \underbrace{\frac1x + \sum_{n=0}^\infty b_nx^n}_{\cot x} - \underbrace{\sum_{n=0}^\infty a_n x^n}_{\tan x} &= 2 \left (\underbrace{\frac{1}{2x} + \sum_{n=0}^\infty b_n(2x)^n}_{\cot 2x} \right) \\
\Rightarrow && \sum_{n=0}^\infty a_n x^n &=  \sum_{n=0}^\infty b_nx^n - 2\sum_{n=0}^\infty b_n(2x)^n \\
&&&= \sum_{n=0}^{\infty}b_n(1-2^{n+1})x^n \\
[x^n]: && a_n &= (1-2^{n+1})b_n
\end{align*}

\item $\,$ \begin{align*}
&& \cot x + \tan x &= \frac{\cos x}{\sin x} + \frac{\sin x}{\cos x} \\
&&&= \frac{1}{\sin x \cos x} \\
&&&=2\cosec 2x \\
\\
\Rightarrow && \underbrace{\frac1x + \sum_{n=0}^\infty b_nx^n}_{\cot x} + \underbrace{\sum_{n=0}^\infty a_n x^n}_{\tan x} &= 2\left (\underbrace{ \frac1{2x} +\sum\limits _{n=0}^\infty c_n (2x)^n}_{\cosec 2x} \right) \\
\Rightarrow && \sum_{n=0}^\infty 2^{n+1}c_n x^n &= \sum_{n=0}^{\infty}(a_n+b_n)x^n \\
&&&= \sum_{n=0}^{\infty}\left((1-2^{n+1})b_n+ b_n\right)x^n \\
&&&= \sum_{n=0}^{\infty}\left(2-2^{n+1}\right)b_nx^n \\
[x^n]: && c_n &= (2^{-n}-1)b_n
\end{align*}

\item $\,$ \begin{align*}
&& \cot^2 x + 1 &= \cosec^2 x \\
\Rightarrow && x^2 \cot^2 x + x^2 &= x^2 \cosec^2 x \\
\Rightarrow && x^2 \left ( \underbrace{\frac1x + \sum_{n=0}^\infty b_nx^n}_{\cot x}  \right)^2 + x^2 &= x^2 \left (\underbrace{ \frac1{x} +\sum\limits _{n=0}^\infty c_n x^n}_{\cosec x}  \right)^2 \\
\Rightarrow &&   \left ( 1 + x\sum_{n=0}^\infty b_nx^{n}  \right)^2 + x^2 &=  \left ( 1 +x\sum\limits _{n=0}^\infty c_n x^{n}  \right)^2 \\
\\
\Rightarrow && \left ( 1 + x(b_1x + b_3 x^3 + \cdots) \right)^2 + x^2 &= \left ( 1 + x(c_1x + c_3 x^3 + \cdots) \right)^2 \\
\Rightarrow && 1 + (1+2b_1)x^2+(2b_3+b_1^2)x^4 + \cdots &= 1 + 2c_1x^2 + (2c_3+c_1^2)x^4 + \cdots \\
\Rightarrow && 1 + 2b_1 &= 2(2^{-1}-1)b_1 \\
\Rightarrow && b_1 &= -\frac13 \\
\Rightarrow && a_1 &= (1-2^{2})(-\tfrac13) = 1 \\
&& c_1 &= \frac16\\
\Rightarrow && 2b_3+\frac19&= 2c_3+\frac1{36} \\
\Rightarrow && 2b_3 -2(2^{-3}-1)b_3 &= -\frac{1}{12} \\
\Rightarrow && \frac{15}{4}b_3 &= -\frac{1}{12} \\
\Rightarrow && b_3 &= -\frac{1}{45} \\
\Rightarrow && a_3 &= -(1-2^4)\frac{1}{45} = \frac13
\end{align*}

\end{questionparts}