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.
 A: \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}
A: Rewrite it as:
$$\frac{\prod_{r=1}^n \left(\cos{\frac{2 \pi}{n}}\sin\frac{(2r-1)\pi}{n}+\sin{\frac{2\pi}{n}}\cos{\frac{(2r-1)\pi}{n}}\right)}{\prod_{r=1}^n \sin{\frac{(2r-1)\pi}{n}}}.$$
The numerator term is $$\cos{\frac{2 \pi}{n}}\sin\frac{(2r-1)\pi}{n}+\sin{\frac{2\pi}{n}}\cos{\frac{(2r-1)\pi}{n}}\\=\sin{\frac{(2r+1)\pi}n}$$
So the numerator and the denominator products are equal, because:
$$\frac{(2r+1)\pi}n=\frac{(2(r+1)-1)\pi}n$$
For $r=1,\dots,n-1$ and:
$$\frac{(2n+1)\pi}n=2\pi+\frac{(2\cdot 1-1)\pi}n$$
