$\displaystyle\sum_{r=1}^{n-1} \cos^2\left ( \dfrac{r\pi}{n} \right )=?$

As we know that

$\cos 2 A=2 \cos ^{2} A-1$

$\cos ^{2} A=\frac{1+\cos 2 A}{2}$

say , $A=\frac{\pi \pi}{n}$

$\Rightarrow \sum_{r=1}^{n-1} 1+\cos 2\left(\frac{r \pi}{n}\right)=\frac{1}{2}\left[\sum_{r=1}^{n-1} \cos \frac{2 \pi \pi}{n}+\sum_{n=1}^{n-1} 1\right]$

$\Rightarrow \frac{1}{2}\left[\cos \frac{2 \pi}{n}+\cos \frac{4 \pi}{n}+\ldots+\cos \frac{2(n-1)}{n} \pi+(n-1)\right]$

As we know that

$\cos \alpha+\cos (\alpha+\beta)+\cos (\alpha+2 \beta)+\cdots+\cos (\alpha+(n-1) \beta)=\frac{\sin \left(\frac{n \beta}{2}\right)}{\sin \frac{\beta}{2}} \cdot \cos \left(\frac{2 \alpha+(n-1) \beta}{2}\right) \\$

$= \frac{1}{2}\left[\frac{\sin \left(\frac{(n-1) \pi}{n}\right)}{\sin \left(\frac{\pi}{n}\right)} \cdot \cos \left(\frac{\frac{4\pi}{n}+(n-2) \frac{2 \pi}{n}}{2}\right)+n-1\right]$

$= \frac{1}{2}\left[\frac{-\sin \left(\frac{n-1}{n} \pi\right)}{\sin \frac{\pi}{n}}+n-1\right]$

$\left.= \frac{1}{2}\left[-\left\{\sin \frac{\sin \left(\pi-\frac{(n-1)}{n}\right) \pi}{\sin \frac{\pi}{n}}\right\}\right\}+n-1\right]$

$= \frac{1}{2}[-\frac{sin\frac{\pi}{n}}{sin\frac{\pi}{n}}+n-1]$

$= \frac{1}{2}(-1+n-1)=\frac{n-2}{2}$

Hence, $(B)$ is the correct option.