[sp]First, $$\begin{aligned}\frac{1}{\cos^210^{\circ}}+\frac{1}{\sin^220^{ \circ }}+\frac{1}{\sin^240^{\circ}}-\frac{1}{\cos^245^{\circ}} &= \Bigl(\frac{1}{\sin^220^{ \circ }}+\frac{1}{\sin^240^{\circ}} +\frac{1}{\sin^260^{\circ}} +\frac{1}{\sin^280^{\circ}}\Bigr) - \frac{1}{\sin^260^{\circ}} - \frac{1}{\cos^245^{\circ}} \\ &= \Bigl(\frac{1}{\sin^220^{ \circ }}+\frac{1}{\sin^240^{\circ}} +\frac{1}{\sin^260^{\circ}} +\frac{1}{\sin^280^{\circ}}\Bigr) - \frac43 -2.\end{aligned}$$ Next, $\theta=20^\circ$, $\theta= 40^\circ$, $\theta= 60^\circ$ and $\theta= 80^\circ$ all satisfy $\sin(9\theta) = 0$. But $\sin(9\theta) = T_9(\sin\theta)$, where $T_9$ is the
Chebyshev polynomial $T_9(x) = 256x^9 - 576x^7 + 432x^5 - 120x^3 + 9x.$ Thus $x = \sin(20k^\circ)\ (k=1,2,3,4)$ are all roots of that polynomial. Dividing by $x$ (because we want to ignore the root $x=0$) and then replacing $x$ by $x^2$, we see that $x = \sin^2(20k^\circ)\ (k=1,2,3,4)$ are the roots of $256x^4 - 576x^3 + 432x^2 - 120x + 9.$ Then replacing $x$ by $1/x$ (and multiplying through by $x^4$), it follows that $x = \dfrac1{\sin^2(20k^\circ)}\ (k=1,2,3,4)$ are the roots of $9x^4 - 120 x^3 + 432 x^2 - 576x^3 + 256 = 0.$ The sum of the roots is $\dfrac{120}9 = \dfrac{40}3.$ Therefore $$\frac{1}{\cos^210^{\circ}}+\frac{1}{\sin^220^{ \circ }}+\frac{1}{\sin^240^{\circ}}-\frac{1}{\cos^245^{\circ}} = \frac{40}3 - \frac43 - 2 = 10.$$[/sp]
