Compute without calculator, $ \frac{1}{\cos^{2}(10)} + \frac{1}{\sin^{2}(20)} + \frac{1}{\sin^{2}(40)} - \frac{1}{\cos^{2}(45)} $ Compute without calculator, 
$$ \frac{1}{\cos^{2}(10^{\circ})} +  \frac{1}{\sin^{2}(20^{\circ})} +   \frac{1}{\sin^{2}(40^{\circ})} -   \frac{1}{\cos^{2}(45^{\circ})} $$

Attempt:
Let $A = \cos(10) \sin(20) \sin(40)$, then
$$ \frac{1}{\cos^{2}(10)} +  \frac{1}{\sin^{2}(20)} +   \frac{1}{\sin^{2}(40)} = \frac{\sin^{2}(20) \sin^{2}(40) + \cos^{2}(10) \sin^{2}(40) + \cos^{2}(10) \sin^{2}(20)}{A^{2}} $$
notice also
$$2A \sin(10) \sin(40) =  \sin^{2}(20) \sin^{2}(40) $$
$$2A \cos(20) \cos(10) =  \cos^{2}(10) \sin^{2}(40) $$
so we have
$$\frac{2A \sin(10) \sin(40) + 2A \cos(20) \cos(10)+ \cos^{2}(10) \sin^{2}(20)}{A^{2}} $$
$$ = \frac{2A \left[2 \sin(10) \sin(20) \cos(20) + \frac{\sqrt{3}}{2} + \sin(10)\sin(20) \right]+ \cos^{2}(10) \sin^{2}(20)}{A^{2}} $$
$$ = \frac{2A \left[\sin(10) \sin(20) (1+ \cos(20) + \cos(20)) + \frac{\sqrt{3}}{2}   \right]+ \cos^{2}(10) \sin^{2}(20)}{A^{2}} $$
$$ = \frac{2A \left[\sin(10) \sin(20) (3\cos^{2}(10) - \sin^{2}(10)) + \frac{\sqrt{3}}{2}   \right]+ \cos^{2}(10) \sin^{2}(20)}{A^{2}} $$
How to continue then?
 A: \begin{align*}
&\;\frac{1}{\cos^{2}10^{\circ}} +  \frac{1}{\sin^{2}20^{\circ}} +   \frac{1}{\sin^{2}40^{\circ}} -   \frac{1}{\cos^{2}45^{\circ}} \\
=&\;\frac {16\sin^210^\circ\cos^220^\circ}{16\sin^210^\circ\cos^210^\circ\cos^220^\circ}+\frac{4\cos^220^\circ}{4\sin^220^\circ\cos^220^\circ}+   \frac{1}{\sin^{2}40^{\circ}} -2\\
=&\;\frac{4(1-\cos20^\circ)(1+\cos40^\circ)+2(1+\cos40^\circ)+1}{\sin^240^\circ}-2\\
=&\;\frac{7-4\cos20^\circ+6\cos40^\circ-4\cos20^\circ\cos40^\circ}{\sin^240^\circ}-2\\
=&\;\frac{7-4\cos20^\circ+6\cos40^\circ-2\cos60^\circ-2\cos20^\circ}{\sin^240^\circ}-2\\
=&\;\frac{6-6\cos20^\circ+6\cos40^\circ}{\sin^240^\circ}-2\\
=&\;\frac{6-12\sin30^\circ\sin 10^\circ}{\sin^240^\circ}-2\\
=&\;\frac{6-6\cos 80^\circ}{\frac12(1-\cos80^\circ)}-2\\
=&\;10
\end{align*}
A: I believe this is how the problem came into being.
Observe that 
$\dfrac1{\cos^210^\circ}=1+\tan^210^\circ$
$\dfrac1{\sin^220^\circ}=\dfrac1{\cos^270^\circ}=1+\tan^270^\circ$
$\dfrac1{\sin^240^\circ}=\dfrac1{\cos^250^\circ}=1+\tan^2(-50^\circ)$
Observe that the angles differ by $60^\circ$
Now if $\tan3x=\tan3A,3x=180^\circ n+3A$ where $n$ is any integer
$x=60^\circ n+A$ where $n=-1,0,1$ or more generally $n\equiv-1,0,1\pmod3$
Again, $\tan3A=\tan3x=\dfrac{3t-t^3}{1-3t^2}$
$\implies t^3-(3\tan3A)t^2-3t+\tan3A=0$ where $t=\tan x$
whose roots are $t_r=\tan(60^\circ r+A)$ where $r=-1,0,1$
$\displaystyle\implies S(A)$
$\displaystyle=\sum_{r=-1}^1t_r^2 =\left(\sum_{r=-1}^1t_r\right)^2-2(t_{-1}t_0+t_0t_1+t_1t_{-1}) =\left(\dfrac{3\tan3A}1\right)^2-2\left(\dfrac{-3}1\right)=6+9\tan^23A$
Here $3A=30^\circ\equiv-150^\circ\equiv210^\circ\pmod{180^\circ}$
So, $\tan3A=\tan30^\circ=\dfrac1{\sqrt3}$
$\implies S(10^\circ)=6+9\left(\dfrac1{\sqrt3}\right)^2=?$
