Showing $\frac{1-\cos \left((n-1)\pi/n\right)}{1-\cos \left(\pi/n\right)} \approx 4n^2/\pi^2$

We have $$\cos \left((n-1)\pi/n\right)=-\cos \left(\pi/n\right)$$ and then $$\frac{1-\cos \left((n-1)\pi/n\right)}{1-\cos \left(\pi/n\right)} =\frac{1+\cos \left(\pi/n\right)}{1-\cos \left(\pi/n\right)}= \frac{2\cos^2 \left(\pi/2n\right)}{2\sin^2 \left(\pi/2n\right)} = \left(\frac{1}{\tan \left(\pi/2n\right)}\right)^2 \approx \left(\frac{1}{\left(\pi/2n\right)}\right)^2=4n^2/\pi^2$$ where $\tan x\approx x$ for small $x$.

To show that $\dfrac{1-\cos ((n-1)\pi/n)}{1-\cos (\pi/n)} \approx 4n^2/\pi^2 $, note that $\cos(2x) =\cos^2(x)-\sin^2(x) =1-2\sin^2(x) $ so $1-\cos(2x) =2\sin^2(x) $ or $1-\cos(x) =2\sin^2(x/2) $.

Therefore

$\begin{array}\\ \dfrac{1-\cos ((n-1)\pi/n)}{1-\cos (\pi/n)} &=\dfrac{2\sin^2((n-1)\pi/(2n))}{2\sin^2(\pi/(2n))} \\ &=\dfrac{\sin^2((1-1/n)\pi/(2))}{\sin^2(\pi/(2n))}\\ &\approx\dfrac{\sin^2(\pi/2-\pi/(2n))}{(\pi/(2n))^2} \qquad\text{since }\sin(x) \approx x \text{ for small } x\\ &=\dfrac{4n^2\cos^2(\pi/(2n))}{\pi^2}\\ &\approx\dfrac{4n^2}{\pi^2} \qquad\text{since } \cos(x) \approx 1 \text{ for small }x\\ \end{array} $