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**Hint:**In this question we will firstly try to reduce the given expression in simpler form then we will check the points where the reduced function can become not defined. Then finally we do not include those points in values of $x$.

**Complete step-by-step answer:**

The given expression is $\left( {1 + \tan x + {{\tan }^2}x} \right)\left( {1 - \cot x + {{\cot }^2}x} \right) \geqslant 3$ -(1)

Let $f\left( x \right) = \left( {1 + \tan x + {{\tan }^2}x} \right)\left( {1 - \cot x + {{\cot }^2}x} \right)$

So now solving the above expression,

$

f\left( x \right) = \left( {1 + \tan x + {{\tan }^2}x} \right)\left( {1 - \cot x + {{\cot }^2}x} \right) \\

{\text{ = 1 - }}\cot x + {\cot ^2}x + \tan x - 1 + \cot x + {\tan ^2}x - \tan x + 1 \\

{\text{ = }}{\tan ^2}x + {\cot ^2}x + 1 \\

$ -(2)

Now using (1) and (2) we can write,

$

f\left( x \right) = {\tan ^2}x + {\cot ^2}x + 1 \geqslant 3 \\

\Rightarrow {\tan ^2}x + {\cot ^2}x \geqslant 2 \\

{\text{ }} \\

$ -(3)

So, here in (3) equation $x \in \mathbb{R}$ satisfy the equation except the points where $\tan x$ and $\cot x$ are not defined. And we know that,

$\tan x$ is not defined when $x \in \dfrac{\pi }{2} + n\pi $, $n \in \mathbb{Z}$

$\cot x$ is not defined when $x \in n\pi $, $n \in \mathbb{Z}$

So, (3) is not defined when,

$

x \in \dfrac{\pi }{2} + n\pi \cup n\pi \\

x \in \left\{ {\dfrac{\pi }{2},\dfrac{{3\pi }}{2},\dfrac{{5\pi }}{2},\dfrac{{7\pi }}{2}, - - - - } \right\} \cup \left\{ {0,\pi ,2\pi ,3\pi , - - - - } \right\} \\

x \in n\pi \\

$ ,$n \in \mathbb{Z}$

Therefore, (1) satisfies when $x \in \mathbb{R}$ excepting $x \in n\pi ,n \in \mathbb{Z}$

**Hence, option (d) is the correct answer.**

**Note:**In the above question when we got (3) equation we can further simplify it into $\sin x$ and $\cos x$. And then we can find the values of x for which equation is not defined. This would be another method to solve this question.

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