If f(x) = cos^2 x + sec^2 x,then | vedclass

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(D) We know that for any real number $a$,the square of a real number is always non-negative,i.e.,$(a - b)^2 \ge 0$.This implies $a^2 + b^2 \ge 2ab$.Le

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$f(x) \ge 2$

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(D) We know that for any real number $a$,the square of a real number is always non-negative,i.e.,$(a - b)^2 \ge 0$.This implies $a^2 + b^2 \ge 2ab$.Let $a = \cos x$ and $b = \sec x = \frac{1}{\cos x}$.Then,$f(x) = \cos^2 x + \sec^2 x = \cos^2 x + \frac{1}{\cos^2 x}$.Using the Arithmetic Mean-Geometric Mean ($AM$-$GM$) inequality,we know that for positive real numbers,the arithmetic mean is greater than or equal to the geometric mean:$\frac{\cos^2 x + \sec^2 x}{2} \ge \sqrt{\cos^2 x \cdot \sec^2 x}$.Since $\cos^2 x \cdot \sec^2 x = \cos^2 x \cdot \frac{1}{\cos^2 x} = 1$,we have:$\frac{f(x)}{2} \ge \sqrt{1} = 1$.Therefore,$f(x) \ge 2$.

If $f(x) = \cos^2 x + \sec^2 x$,then

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