For non negative integer $n$, suppose that  $f\left(n\right)=\frac{{\displaystyle \sum _{k=0}^n\sin \left(\frac{k+1}{n+2}\pi \right)\sin \left(\frac{k+2}{n+2}\pi \right)}}{{\displaystyle \sum _{k=0}^n{\sin }^2\left(\frac{k+1}{n+2}\pi \right)}}$, Assuming ${\cos }^{-1}\left(x\right)$ takes values in $\left[0,\pi \right]$

 which of the following options is/are correct?

$f\left(n\right)=\frac{{\displaystyle \sum _{k=0}^n\left(\cos \left(\frac{\pi }{n+2}\right)-\cos \left(\frac{2k+3}{n+2}\pi \right)\right)}}{{\displaystyle \sum _{k=0}^n\left(1-\cos \left(\frac{2k+2}{n+2}\pi \right)\right)}}$

$\begin{array}{l}=\frac{\left(n+1\right)\cos \left(\frac{\pi }{n+2}\right)-\frac{\sin \left(\frac{(n+1)\pi }{n+2}\right)}{\sin \left(\frac{\pi }{n+2}\right)}\cos \left(\frac{3\pi }{n+2}+\frac{n\pi }{n+2}\right)}{\left(n+1\right)-\frac{\sin \left(\frac{(n+1)\pi }{n+2}\right)}{\sin \left(\frac{\pi }{n+2}\right)}\cos \left(\frac{2\pi }{n+2}+\frac{n\pi }{n+2}\right)}\mid \\ =\frac{\left(n+2\right)\cos \left(\frac{\pi }{n+2}\right)}{n+2}=\cos \left(\frac{\pi }{n+2}\right).\\ \therefore f\left(6\right)=\cos \left(\frac{\pi }{8}\right)\Rightarrow \alpha =Tan\frac{\pi }{8}=\sqrt{2}-1\Rightarrow {\alpha }^2+2\alpha -1=0\end{array}$