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**Hint:**In this question, first of fid the maximum and maximum of \[{\cos ^2}\theta \] and \[{\sec ^2}\theta \] by using the concept that the values of \[\cos \theta \] will always lies between \[\left[ { - 1,1} \right]\] and the values of \[\sec \theta \] will always between \[\left( { - \infty , - 1} \right) \cup \left( {1,\infty } \right)\]. Then find out the maximum value of the given function to reach the solution of the given problem.

__Complete step-by-step answer__:Given that \[f\left( x \right) = {\cos ^2}x + {\sec ^2}x\]

We know that the values of \[\cos \theta \] will always lies between \[\left[ { - 1,1} \right]\]

So, the values of \[{\cos ^2}\theta \] lies between \[\left[ {0,1} \right]\]

Hence, the maximum value of \[{\cos ^2}\theta \] is 1 and the minimum value of \[{\cos ^2}\theta \] is 0.

Also, we know that the values of \[\sec \theta \] will always between \[\left( { - \infty , - 1} \right) \cup \left( {1,\infty } \right)\]

So, the values of \[{\sec ^2}\theta \] lies between \[\left( {1,\infty } \right)\]

Hence, the maximum value of \[{\sec ^2}\theta \] is \[\infty \] and the minimum value of \[{\sec ^2}\theta \] is 1.

Since \[{\cos ^2}\theta = \dfrac{1}{{{{\sec }^2}\theta }}\], when \[{\cos ^2}\theta \] has its minima \[{\sec ^2}\theta \] has its maxima and vice-versa.

Let us consider the maximum value of the given function \[f\left( x \right)\]

\[

\Rightarrow f{\left( x \right)_{\max }} = \max {\cos ^2}\theta + \min {\sec ^2}\theta \\

\Rightarrow f{\left( x \right)_{\max }} = 1 + 1 \\

\Rightarrow f{\left( x \right)_{\max }} = 2 \\

\]

Therefore, \[f\left( x \right) \geqslant 2\].

**Thus, the correct option is D. \[f\left( x \right) \geqslant 2\]**

**Note**: As \[{\cos ^2}\theta = \dfrac{1}{{{{\sec }^2}\theta }}\], when \[{\cos ^2}\theta \] has its minima \[{\sec ^2}\theta \] has its maxima and when \[{\cos ^2}\theta \] has its maxima \[{\sec ^2}\theta \] has its minima. Always remember the range and domain of various trigonometric functions to solve these kinds of questions.

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