Random Trigonometry that we do not need in daily life

Consider the double-angle formula

$sin \space 2A = 2sin \space A \space cos \space A$

$cos \space 2A = 2cos^{2} A - 1$

We have

$\frac{1}{cos^{2} 10^{\circ}} = \frac{1}{sin^{2} 80^{\circ}}$

$\frac{1}{sin^{2} 20^{\circ}} = \frac{4 cos^{2} 20^{\circ}}{sin^{2} 40^{\circ}} = \frac{2(1+cos \space 40^{\circ})}{sin^{2} 40^{\circ}} = \frac{2(1+cos \space 40^{\circ}) \space\cdot\space 4 cos^{2} 40^{\circ}}{sin^{2} 80^{\circ}}$

$\frac{1}{sin^{2} 40^{\circ}} = \frac{4 cos^{2} 40^{\circ}}{sin^{2} 80^{\circ}}$

Then

$\frac{1}{cos^{2} 10^{\circ}} + \frac{1}{sin^{2} 20^{\circ}} + \frac{1}{sin^{2} 40^{\circ}} = \frac{1+(3+2cos \space 40^{\circ}) \space\cdot\space 4 cos^{2} 40^{\circ}}{cos^{2} 10^{\circ}}$

Now consider the numerator

$1+(3+2cos \space 40^{\circ}) \cdot 4 cos^{2} 40^{\circ}$

$= 1+(3+2cos \space 40^{\circ}) \cdot 2(1+ cos \space 80^{\circ})$

$= 1+(6+4cos \space 40^{\circ}+6cos \space 80^{\circ}+4cos \space 40^{\circ}cos \space 80^{\circ})$

$= 1+6+4cos \space 40^{\circ}+6cos \space 80^{\circ}+2(cos \space 120^{\circ}+cos \space 40^{\circ})$

$= 6+6cos \space 40^{\circ}+6cos \space 80^{\circ}$

$= 6+6 \cdot (2cos \space 60^{\circ}cos \space 20^{\circ})$

$= 6(1+cos \space 20^{\circ})$

$= 12cos^{2} 10^{\circ}$

Therefore,

$\frac{1}{cos^{2} 10^{\circ}} + \frac{1}{sin^{2} 20^{\circ}} + \frac{1}{sin^{2} 40^{\circ}} - \frac{1}{cos^{2} 45^{\circ}} = 12-2 = 10$