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Consider $\displaystyle{S = \sum_{k = 1}^{n - 1}\csc\left(k\,{\pi \over n}\right) = \frac{1}{\sin\left(\pi/n\right)} + \frac{1}{\sin\left(2\pi/n\right)} + \frac{1}{\sin\left(3\pi/n\right)} + \cdots + \frac{1}{\sin\left(\left[n - 1\right]\pi/n\right)}}$

How can we find a general formula for $S$ using trigonometry identities or complex numbers ?.

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Let we assume for first that $n$ is an odd number, $n=2N+1$. In such a case the given sum is $$ \sum_{k=1}^{2N}\frac{1}{\sin\left(\frac{\pi k}{2N+1}\right)} = 2\sum_{k=1}^{N}\frac{1}{\sin\left(\frac{\pi k}{2N+1}\right)}=\frac{4N+2}{\pi}H_N+2\sum_{k=1}^{N}\left[\frac{1}{\sin\left(\frac{\pi k}{2N+1}\right)}-\frac{1}{\frac{\pi k}{2N+1}}\right] $$ and $\frac{1}{\sin(x)}-\frac{1}{x}$ is an integrable function on the interval $\left(0,\frac{\pi}{2}\right)$, whose integral equals $\log\frac{4}{\pi}$.$^{(*)}$
By Riemann sums is follows that: $$ \sum_{k=1}^{2N}\frac{1}{\sin\left(\frac{\pi k}{2N+1}\right)} = \frac{4N}{\pi}\left[H_N+\log\frac{4}{\pi}\right]+O(\log N).$$ In the general case we get that the given sum behaves like $Cn\log n$.

$^{(*)}$ Since $\text{Res}\left(\frac{1}{\sin x},x=k\pi\right)=(-1)^k$, by Herglotz' trick we have $$\frac{1}{\sin x}-\frac{1}{x}=\sum_{k\geq 1}\left(\frac{1}{x-k\pi}+\frac{1}{x+k\pi}\right)(-1)^k $$ and by termwise integration $$ \int_{0}^{\pi/2}\left(\frac{1}{\sin x}-\frac{1}{x}\right)\,dx = \sum_{k\geq 1}(-1)^k \log\left(1-\frac{1}{4k^2}\right) $$ so $I=\log\frac{4}{\pi}$ by simplifying the partial sums of the last series and recalling Wallis product.

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    Does your sum not blow up? – 2017-01-18
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    @RonGordon: the given sum behaves like $n\log n$ hence yes, it does blow up. – 2017-01-18
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    No, I mean actually blow up. Look at the first sum when $k=2 N$. – 2017-01-18
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    @RonGordon: oh, sorry, wrong summation indices. Now fixing. – 2017-01-18
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    @JackD'Aurizio Can you explain about $H_N$ ? – 2017-01-18
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    @S.H.W: $H_N$ is the $N$-th harmonic number, defined by $$H_N = \sum_{k=1}^{N}\frac{1}{k} $$ and behaving like $\log(N)+\gamma$. – 2017-01-18
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    I just separated the "singular part" from the "regular part" and estimated them both with decent accuracy. – 2017-01-18
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    @JackD'Aurizio What about $O(log N)$ ? Is it approximation ? – 2017-01-18
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    @S.H.W: well, $O(\log N)$ means $O(\log N)$. Such term comes from neglecting the contribute given by $\frac{2}{\pi}H_N$ in the first line. – 2017-01-18
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    @JackD'Aurizio Your answer is very good and fantastic but do you know any more simple answer using trigonometry or complex numbers ? – 2017-01-18
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    @S.H.W it won't get any simpler – 2017-01-18
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    @S.H.W: the behaviour of your sum depends (almost only) on the singular part, that is not a trigonometric series, it is just a partial sum for the harmonic series. Despite its appearance, this question is not really related with trigonometric/complex number identities, it is about understanding the consequences of a non-integrable singularity for a positive real function. – 2017-01-18
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    Okay , Thanks a lot for your help. – 2017-01-18