Simplify the product $\prod_{r=1}^n\left(\cos{\frac{2 \pi}{n}}+\sin{\frac{2\pi}{n}}\cdot \cot{\frac{(2r-1)\pi}{n}}\right)$, where $n$ is even Question: Simplify the product as far as possible: $$\prod_{r=1}^n\left(\cos{\frac{2 \pi}{n}} + \sin{\frac{2\pi}{n}}\cdot \cot{\frac{(2r-1)\pi}{n}}\right),$$ where $n$ is even.
The question is taken from a University entrance exam from the UK called the STEP, for which I am unable to find the solution to online.
When attempting similar questions my approach was to substitute values in order to spot a pattern however this isn't really feasible in this question.

\begin{align}
&\prod_{r=1}^n\left(\cos{\frac{2 \pi}{n}} + \sin{\frac{2\pi}{n}}\cdot \cot{\frac{(2r-1)\pi}{n}}\right)=\\
&=\prod_{r=1}^n\frac{\cos{\frac{2 \pi}{n}}\sin{\frac{(2r-1)\pi}{n}} + \sin{\frac{2\pi}{n}}\cdot \cos{\frac{(2r-1)\pi}{n}}}{\sin{\frac{(2r-1)\pi}{n}}}\\
&=\prod_{r=1}^n\frac{\sin\left({\frac{2 \pi}{n}}+{\frac{(2r-1)\pi}{n}}\right)}{\sin{\frac{(2r-1)\pi}{n}}}\\
&=\prod_{r=1}^n\frac{\sin{\frac{(2r+1)\pi}{n}}}{\sin{\frac{(2r-1)\pi}{n}}}\\
&=\frac{\sin{\frac{(2n+1)\pi}{n}}}{\sin{\frac{\pi}{n}}}\\
&=\frac{\sin{\frac{\pi}{n}}}{\sin{\frac{\pi}{n}}}\\
&=1
\end{align}
