$Foralln\ge 1,let$

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

By the double angle and sum-product identities for cosine, we have
$2{\cos }^2\left(\frac{(k-1)\pi }{2n}\right)-2{\cos }^2\left(\frac{k\pi }{2n}\right)=\cos \left(\frac{(k-1)\pi }{n}\right)-\cos \left(\frac{k\pi }{n}\right)$
$=2\sin \left(\frac{(2k-1)\pi }{2n}\right)\sin \left(\frac{\pi }{2n}\right)$
and it follows that the summand in  $f\left(n\right)$ can be written as 
$\frac{1}{\sin \left(\frac{\pi }{2n}\right)}(-\frac{1}{{\cos }^2\left(\frac{(k-1)\pi }{2n}\right)}+\frac{1}{{\cos }^2\left(\frac{k\pi }{2n}\right)})$.
Thus the sum telescopes and we find that
$f\left(n\right)=\frac{1}{\sin \left(\frac{\pi }{2n}\right)}(-1+\frac{1}{{\cos }^2\left(\frac{(n-1)\pi }{2n}\right)})=-\frac{1}{\sin \left(\frac{\pi }{2n}\right)}+\frac{1}{{\sin }^3\left(\frac{\pi }{2n}\right)}$.
Finally, since $\underset{x\to 0}{\lim }\frac{\sin x}{x}=1,wehave$
 ${\lim }_{n\to \infty }\left(n\sin \frac{\pi }{2n}\right)=\frac{\pi }{2},andthus{\lim }_{n\to \infty }\frac{f\left(n\right)}{n^3}=\frac{8}{{\pi }^3}.$