Problem source

Show that 
\[
\sin(2n+1)\theta=\sin^{2n+1}\theta\sum_{r=0}^{n}(-1)^{n-r}\binom{2n+1}{2r}\cot^{2r}\theta,
\]
where $n$ is a positive integer. Deduce that the equation 
\[
\sum_{r=0}^{n}(-1)^{r}\binom{2n+1}{2r}x^{r}=0
\]
has roots $\cot^{2}(k\pi/(2n+1))$ for $k=1,2,\ldots,n$. 

Show that 

\begin{itemize}
\setlength{\itemsep}{3mm}
\item[\bf (i)]  ${\displaystyle \sum_{k=1}^{n}\cot^{2}\left(\frac{k\pi}{2n+1}\right)=\frac{n(2n-1)}{3}},$ 

\item[\bf (ii)] ${\displaystyle \sum_{k=1}^{n}\tan^{2}\left(\frac{k\pi}{2n+1}\right)=n(2n+1)},$ 

\item[\bf (iii)] ${\displaystyle \sum_{k=1}^{n}\mathrm{cosec}^{2}\left(\frac{k\pi}{2n+1}\right)=\frac{2n(n+1)}{3}}.$
\end{itemize}