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**Hint:**we can solve this problem by the simplifying given term.

Here we can use some trigonometric formulas.

1) $\sin 2\theta = 2\sin \theta \cos \theta $

2) $\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B$

3) $\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$

4) ${\sin ^2}\theta + {\cos ^2}\theta = 1$

**Complete step by step answer:**

1) Let us simplify the given term.

We have,

$ = \sqrt {4{{\cos }^4}\theta + {{\sin }^2}2\theta } + 4\cot \theta {\cos ^2}\left( {\dfrac{\pi }{4} - \dfrac{\theta }{2}} \right)$

Let us simplify first the terms in the square root.

We can use 1) formula.

We get,

$ \Rightarrow \sqrt {4{{\cos }^4}\theta + {{\left( {2\sin \theta \cos \theta } \right)}^2}} + 4\cot \theta {\cos ^2}\left( {\dfrac{\pi }{4} - \dfrac{\theta }{2}} \right)$

Simplify the remaining term.

Let us use 2) formula.

Now we get,

$ \Rightarrow \sqrt {4{{\cos }^4}\theta + {{\left( {2\sin \theta \cos \theta } \right)}^2}} + 4\cot \theta {\left[ {\cos \dfrac{\pi }{4}\cos \dfrac{\theta }{2} + \sin \dfrac{\pi }{4}\sin \dfrac{\theta }{2}} \right]^2}$

Now, we can put the values of cos and sin.

As we know the values of, $\cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$ and$\sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$.

Substitute these values.

$ \Rightarrow \sqrt {4{{\cos }^4}\theta + {{\left( {2\sin \theta \cos \theta } \right)}^2}} + 4\cot \theta {\left[ {\dfrac{1}{{\sqrt 2 }}\cos \dfrac{\theta }{2} + \dfrac{1}{{\sqrt 2 }}\sin \dfrac{\theta }{2}} \right]^2}$

Consider the term which is under the square root and is in the bracket.

Take the square of that term.

See, $\dfrac{1}{{\sqrt 2 }}$ is in both the terms.

So, let us write it outside the bracket as a common term.

So we get,

$ \Rightarrow \sqrt {4{{\cos }^4}\theta + \left( {4{{\sin }^2}\theta {{\cos }^2}\theta } \right)} + \dfrac{4}{{{{\left( {\sqrt 2 } \right)}^2}}}\cot \theta {\left[ {\cos \left( {\dfrac{\theta }{2}} \right) + \sin \left( {\dfrac{\theta }{2}} \right)} \right]^2}$

Consider the term which is under the square root.

Let us take common terms outside the bracket.

So we get,

$ \Rightarrow \sqrt {4{{\cos }^2}\theta \left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right)} + \dfrac{4}{{{{\left( {\sqrt 2 } \right)}^2}}}\cot \theta {\left[ {\cos \left( {\dfrac{\theta }{2}} \right) + \sin \left( {\dfrac{\theta }{2}} \right)} \right]^2}$

While removing the square root so, the value which is in the square root will be in the mod.

After that, we will use the formula to simplify the bracket:${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$.

$ \Rightarrow \left| {2\cos \theta \left( 1 \right)} \right| + \dfrac{4}{2}\cot \theta \left[ {{{\cos }^2}\left( {\dfrac{\theta }{2}} \right) + {{\sin }^2}\left( {\dfrac{\theta }{2}} \right) + 2\sin \left( {\dfrac{\theta }{2}} \right)\cos \left( {\dfrac{\theta }{2}} \right)} \right]$

Use the fourth formula which is in the hint.

$ \Rightarrow \left| {2\cos \theta } \right| + 2\cot \theta \left[ {1 + \sin \theta } \right]$

Simplify the bracket.

$ \Rightarrow \left| {2\cos \theta } \right| + 2\cot \theta + 2\cot \theta \sin \theta $

As we know, $\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$.

$ \Rightarrow \left| {2\cos \theta } \right| + 2\cot \theta + 2\dfrac{{\cos \theta }}{{\sin \theta }}\sin \theta $

Here, $\sin \theta $ get cancelled.

$ \Rightarrow 2\left[ {\left| {\cos \theta } \right| + \cot \theta + \cos \theta } \right]$

In both the 2 & 3 quadrants we have given the value of cos is negative.

As the first cos is in the mod. So, there is no need to change the sign of that term.

$ \Rightarrow 2\left[ {\left| {\cos \theta } \right| + \cot \theta - \cos \theta } \right]$

$ \Rightarrow 2\left[ {\cot \theta } \right]$

$ \Rightarrow 2\cot \theta $

**Therefore, option (C) is the correct answer.**

**Note:**

- To solve this kind of problem, students should be familiar with trigonometric basic formulas, identities, reciprocal identities, angle sum & difference identities.

- Students should know about trigonometric functions and the signs of main trigonometric functions sin, cos, tan in different quadrants.

- Important thing is that students should know the values of all trigonometric functions at various degrees.

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