Problem based on sum of reciprocal of $n^{th}$ roots of unity

Let $1,x_{1},x_{2},x_{3},\ldots,x_{n-1}$ be the $\bf{n^{th}}$ roots of unity. Find: $$\frac{1}{1-x_{1}}+\frac{1}{1-x_{2}}+......+\frac{1}{1-x_{n-1}}$$

$\bf{My\; Try::}$ Given $x=(1)^{\frac{1}{n}}\Rightarrow x^n=1\Rightarrow x^n-1=0$

Now Put $\displaystyle y = \frac{1}{1-x}\Rightarrow 1-x=\frac{1}{y}\Rightarrow x=\frac{y-1}{y}$

Put that value into $\displaystyle x^n-1=0\;,$ We get $\displaystyle \left(\frac{y-1}{y}\right)^n-1=0$

So we get $(y-1)^n-y^n=0\Rightarrow \displaystyle \left\{y^n-\binom{n}{1}y^{n-1}+\binom{n}{2}y^2-......\right\}-y^n=0$

So $\displaystyle \binom{n}{1}y^{n-1}-\binom{n}{2}y^{n-1}+...+(-1)^n=0$ has roots $y=y_{1}\;,y_{2}\;,y_{3}\;,.......,y_{n-1}$

So $$y_{1}+y_{2}+y_{3}+.....+y_{n-1} = \binom{n}{2} =\frac{n(n-1)}{2}\;,$$ Where $\displaystyle y_{i} = \frac{1}{1-x_{i}}\;\forall i\in \left\{1,2,3,4,5,.....,n-1\right\}$

My Question is can we solve it any less complex way, If yes Then plz explain here, Thanks

Let $f(x) = \frac{(x^n-1)}{(x-1)} = x^{n-1}+x^{n-2}+...+x+1$

Then $\frac{f'(x)}{f(x)} = \frac{1}{x-x_1} + \frac{1}{x-x_2}+ ... + \frac{1}{x-x_{n-1}}$

So what you want is just $\frac{f'(1)}{f(1)}$, which is easy to compute as $\frac{n-1}{2}$.

Otherwise you could write $x_i=e^{j\frac {i2\pi} {n}}$ ($j^2=-1$)

Then $$y_i=\frac {e^{-j\frac {i\pi} {n}}} {e^{-j\frac {i\pi} {n}}-e^{j\frac {i\pi} {n}}}=\frac {e^{-j\frac {i\pi} {n}}}{-2j\sin(\frac {i\pi} {n})}=2j\cot(\frac {i\pi} {n})+\frac 1 2$$

Now consider the fact that, if $x_i$ is an $n^{th}$ root of unity, so is $\bar{x}_i$, so if $x_i\neq \bar x_i$ , your sum contains $y_i+\bar y_i=2\mathcal {Re}(y_i)=1$ exactly $\frac {n-1} {2}$ if $n$ is odd or $\frac {n-2} {2}$ if $n$ is even (but the remaining term is $\frac 1 {1-(-1)}=\frac 1 2$).

So in both cases the result is $\frac {n-1} 2$

I don't know if this can be considered "less complex" but it certainly involves more "down to earth" calculations IMHO.