Question:
If $f(x)=\cos ^{2} x+\sec ^{2} x$, then
(a) f(x) < 1
(b) f(x) = 1
(c) 1 < f(x) < 2
(d) f(x) ≥ 2
Solution:
$f(x)=\cos ^{2} x+\sec ^{2} x$
$=\cos ^{2} x+\sec ^{2} x-2 \cos x \sec x+2 \cos x \sec x$
$=(\sec x-\cos x)^{2}+2$
$\therefore f(x) \geq 2 \forall x \quad\left[(\sec x-\cos x)^{2} \geq 0 \forall x\right]$
Hence, the correct option is answer D.