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

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(D) Given the function $f(x) = \cos^2 x + \sec^2 x$.We can rewrite the expression by completing the square:$f(x) = (\cos x - \sec x)^2 + 2 \cos x \sec

Vedclass · Accepted Answer

$f(x) \geq 2$

Vedclass · Answer

(D) Given the function $f(x) = \cos^2 x + \sec^2 x$.We can rewrite the expression by completing the square:$f(x) = (\cos x - \sec x)^2 + 2 \cos x \sec x$Since $\cos x \cdot \sec x = 1$,the expression becomes:$f(x) = (\cos x - \sec x)^2 + 2(1)$$f(x) = (\cos x - \sec x)^2 + 2$Since the square of any real number is always non-negative,i.e.,$(\cos x - \sec x)^2 \geq 0$,it follows that:$f(x) \geq 0 + 2$$f(x) \geq 2$Therefore,the value of the function is always greater than or equal to $2$.

If $f(x) = \cos^2 x + \sec^2 x$,then its value is always:

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