Let's boycott Calculus this time! (Part 2)

Just use Algebra, AM-GM inequality :

$\begin{aligned} \cos^2 x+\sec^2 x \ge 2\sqrt{\cos^2 x \sec ^2 x} =\boxed {2}\end{aligned}$

Equality occurs when $\cos x=\sec x=1$ .

$= \cos^2(x) + \sec^2(x)$

$= \cos^2(x) + \frac{1}{\cos^2(x)}$

$= \frac{1+\cos^4(x)}{\cos^2(x)}$

$= \frac{[1-\cos^2(x)]^2 + 2\cos^2(x)}{\cos^2(x)}$

$= [\frac{\sin^2(x)}{\cos(x)}]^2 + 2$

Since $\frac{\sin^2(x)}{\cos(x)}$ is a real value for $x \neq (2n+1)\frac{π}{2}$ , $[\frac{\sin^2(x)}{\cos(x)}]^2$ is a positive real number or $0$ . Clearly then, we have $[\frac{\sin^2(x)}{\cos(x)}]^2 + 2 \geqslant \boxed{2}$