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different approach.
we need to determine
\begin{align}
\lim_{n\to\infty} \frac{1}{n^3}\sum_{k=1}^{n-1}\frac{\sin\left(\frac{(2k-1)\pi}{2n}\right)}{\cos^{2}\left(\frac{(k-1)\pi}{2n}\right)\cos^2\left(\frac{k\pi}{2n}\right)}
&=\lim_{n\to\infty} \frac{1}{n^3}\sum_{k=1}^{n-1}\frac{\sin\left(\frac{k\pi}{2n}+\frac{(k-1)\pi}{2n}\right)}{\cos^{2}\left(\frac{(k-1)\pi}{2n}\right)\cos^2\left(\frac{k\pi}{2n}\right)}
\\&=\lim_{n\to\infty} \frac{1}{n^3}\sum_{k=1}^{n-1}\frac{\sin\left(\frac{k\pi}{2n}\right)\cos\left(\frac{(k-1)\pi}{2n}\right)+\sin\left(\frac{(k-1)\pi}{2n}\right)\cos\left(\frac{k\pi}{2n}\right)}{\cos^{2}\left(\frac{(k-1)\pi}{2n}\right)\cos^2\left(\frac{k\pi}{2n}\right)}
\\ &= \lim_{n\to\infty} \frac{1}{n^3}\sum_{k=1}^{n-1}\frac{\tan\left(\frac{k\pi}{2n}\right)+\tan\left(\frac{(k-1)\pi}{2n}\right)}{\cos\left(\frac{(k-1)\pi}{2n}\right)\cos\left(\frac{k\pi}{2n}\right)}
\\ &= \lim_{n\to\infty} \frac{1}{n^3}\sum_{k=1}^{n-1}\frac{\left(\tan\left(\frac{k\pi}{2n}\right)+\tan\left(\frac{(k-1)\pi}{2n}\right)\right)\left(\tan\left(\frac{k\pi}{2n}\right)-\tan\left(\frac{(k-1)\pi}{2n}\right)\right)}{\cos\left(\frac{(k-1)\pi}{2n}\right)\cos\left(\frac{k\pi}{2n}\right)\left(\tan\left(\frac{k\pi}{2n}\right)-\tan\left(\frac{(k-1)\pi}{2n}\right)\right)}
\\ &= \lim_{n\to\infty} \frac{1}{n^3}\sum_{k=1}^{n-1}\frac{\tan^2\left(\frac{k\pi}{2n}\right)-\tan^2\left(\frac{(k-1)\pi}{2n}\right)}{\sin\left(\frac{k\pi}{2n}\right)\cos\left(\frac{(k-1)\pi}{2n}\right)-\sin\left(\frac{(k-1)\pi}{2n}\right)\cos\left(\frac{k\pi}{2n}\right)}
\\ &= \lim_{n\to\infty} \frac{1}{n^3}\sum_{k=1}^{n-1}\frac{\tan^2\left(\frac{k\pi}{2n}\right)-\tan^2\left(\frac{(k-1)\pi}{2n}\right)}{\sin\left(\frac{k\pi}{2n}-\frac{(k-1)\pi}{2n}\right)}
\\ &= \lim_{n\to\infty} \frac{\csc\left(\frac{\pi}{2n}\right)}{n^3}\sum_{k=1}^{n-1}\tan^2\left(\frac{k\pi}{2n}\right)-\tan^2\left(\frac{(k-1)\pi}{2n}\right)
\\ &= \lim_{n\to\infty} \frac{\csc\left(\frac{\pi}{2n}\right)}{n^3}\left(\sum_{k=2}^{n}\tan^2\left(\frac{(k-1)\pi}{2n}\right)-\sum_{k=1}^{n-1}\tan^2\left(\frac{(k-1)\pi}{2n}\right)\right)
\\ &= \lim_{n\to\infty} \frac{\csc\left(\frac{\pi}{2n}\right)}{n^3}\left(\tan^2\left(\frac{(n-1)\pi}{2n}\right)-\tan^20\right)
\\ &= \lim_{n\to\infty} \frac{\tan^2\left(\frac{(n-1)\pi}{2n}\right)}{n^3\sin\left(\frac{\pi}{2n}\right)} \ \ \ \ \ (*_1)
\\ &= \lim_{n\to\infty} \frac{\tan^2\left(\frac \pi 2 -\frac{\pi}{2n}\right)}{n^3\left(\frac{\pi}{2n}\right)}
\\ &= \frac 2 \pi \lim_{n\to\infty} \frac{\cot^2\left(\frac{\pi}{2n}\right)}{n^2}
\\ &= \frac 2 \pi\lim_{n\to\infty} \frac{\cot\left(\frac{\pi}{2n}\right)\csc^2\left(\frac{\pi}{2n}\right)\left(\frac{\pi}{n^2}\right)}{2n} \ \ \ \ \ (*_2)
\\ &= \lim_{n\to\infty} \frac{\cot\left(\frac{\pi}{2n}\right)\csc^2\left(\frac{\pi}{n}\right)}{n^3}
\\ &= \lim_{n\to\infty} \frac{\cos\left(\frac{\pi}{2n}\right)}{\left(n\sin\left(\frac{\pi}{2n}\right)\right)^3}
\\ &= \lim_{n\to\infty} \frac{1-\frac 1 2 \left(\frac{\pi}{2n}\right)^2}{\left(n\left(\frac{\pi}{2n}\right)\right)^3} \ \ \ \ \ (*_3),(*_4)
\\ &= \frac{1}{\left(\frac{\pi}{2}\right)^3}
\\ &=\frac{8}{\pi^3}
\end{align}
noting that (*_1),(*_3),(*_4) are true due to small-angle approximations, and (*_2) is true due to l’hôpital’s rule
well that was quite \sqrt{1+\tan^2c} (get it)
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