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

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(D) We know that for any real number $a$,$(a - \frac{1}{a})^2 \ge 0$.This implies $a^2 + \frac{1}{a^2} - 2 \ge 0$,or $a^2 + \frac{1}{a^2} \ge 2$.Let $

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$f(x) \ge 2$

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(D) We know that for any real number $a$,$(a - \frac{1}{a})^2 \ge 0$.This implies $a^2 + \frac{1}{a^2} - 2 \ge 0$,or $a^2 + \frac{1}{a^2} \ge 2$.Let $a = \cos x$. Since $\sec x = \frac{1}{\cos x}$,we have $f(x) = \cos^2 x + \frac{1}{\cos^2 x}$.By the Arithmetic Mean-Geometric Mean inequality ($AM$ $\ge$ $GM$),for positive real numbers $u$ and $v$,$\frac{u+v}{2} \ge \sqrt{uv}$.Here,$f(x) = \cos^2 x + \sec^2 x \ge 2 \sqrt{\cos^2 x \cdot \sec^2 x} = 2 \sqrt{1} = 2$.Thus,$f(x) \ge 2$.

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

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