We can write \begin{align*} \frac{1}{\cos^2 10^\circ} &= \frac{2}{1 + \cos 20^\circ} \\ &= \frac{2 (1 - \cos 20^\circ)}{(1 + \cos 20^\circ)(1 - \cos 20^\circ)} \\ &= \frac{2 (1 - \cos 20^\circ)}{1 - \cos^2 20^\circ} \\ &= \frac{2 - 2 \cos 20^\circ}{\sin^2 20^\circ}, \end{align*}so \begin{align*} \frac{1}{\cos^2 10^\circ} + \frac{1}{\sin^2 20^\circ} + \frac{1}{\sin^2 40^\circ} &= \frac{2 - 2 \cos 20^\circ}{\sin^2 20^\circ} + \frac{1}{\sin^2 20^\circ} + \frac{1}{\sin^2 40^\circ} \\ &= \frac{3 - 2 \cos 20^\circ}{\sin^2 20^\circ} + \frac{1}{\sin^2 40^\circ} \\ &= \frac{4 \cos^2 20^\circ (3 - 2 \cos 20^\circ)}{4 \sin^2 20^\circ \cos^2 20^\circ} + \frac{1}{\sin^2 40^\circ} \\ &= \frac{12 \cos^2 20^\circ - 8 \cos^3 20^\circ}{\sin^2 40^\circ} + \frac{1}{\sin^2 40^\circ} \\ &= \frac{12 \cos^2 20^\circ - 8 \cos^3 20^\circ + 1}{\sin^2 40^\circ}.\end{align*} By the triple angle formula, \begin{align*} \frac{1}{2} &= \cos 60^\circ \\ &= \cos (3 \cdot 20^\circ) \\ &= 4 \cos^3 20^\circ - 3 \cos 20^\circ, \end{align*}which means $ 8 \cos^3 20^\circ = 6 \cos 20^\circ + 1 $. Hence, \begin{align*} \frac{12 \cos^2 20^\circ - 8 \cos^3 20^\circ + 1}{\sin^2 40^\circ} &= \frac{12 \cos^2 20^\circ - 6 \cos 20^\circ}{\sin^2 40^\circ} \\ &= \frac{12 \cos^2 20^\circ - 6 \cos 20^\circ}{4 \sin^2 20^\circ \cos^2 20^\circ} \\ &= \frac{12 \cos 20^\circ - 6}{4 \sin^2 20^\circ \cos 20^\circ} \\ &= \frac{12 \cos 20^\circ - 6}{4 (1 - \cos^2 20^\circ) \cos 20^\circ} \\ &= \frac{12 \cos 20^\circ - 6}{4 \cos 20^\circ - 4 \cos^3 20^\circ} \\ &= \frac{12 \cos 20^\circ - 6}{4 \cos 20^\circ - 3 \cos 20^\circ - \frac{1}{2}} \\ &= \frac{12 \cos 20^\circ - 6}{\cos 20^\circ - \frac{1}{2}} \\ &= \boxed{12}.\end{align*}