What is ${\cot ^2}\theta $. in terms of non- exponential trigonometric function?

Hint: In this question, we are given a trigonometric function ${\cot ^2}\theta $. And we have to convert it in non-exponential trigonometric function i.e., we have to make its degree one.
For that, we will first write $\cot \theta $ in the form of $\sin \theta $ and $\cos \theta $ .
Then, we will use the half-angle formulas for removing their exponential powers.
Formulae to be used:
$\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$ ,
$\cos 2\theta = 1 - 2{\sin ^2}\theta $ ,
$\cos 2\theta = 2{\cos ^2}\theta - 1$ .

Complete answer:
Given trigonometric function is ${\cot ^2}\theta $ .
To write the given trigonometric function in terms of the non-exponential function.
For that, first, we will write $\cot \theta $ in the form of $\sin \theta $ and $\cos \theta $ , i.e., we can write it as \[{\cot ^2}\theta = \dfrac{{{{\cos }^2}\theta }}{{{{\sin }^2}\theta }}\] .
Now, we know that, $\cos 2\theta = 1 - 2{\sin ^2}\theta $ , so, adding $2{\sin ^2}\theta $ on both sides, we get, $\cos 2\theta + 2{\sin ^2}\theta = 1 - 2{\sin ^2}\theta + 2{\sin ^2}\theta $ , i.e., $\cos 2\theta + 2{\sin ^2}\theta = 1$ . Now, subtracting $\cos 2\theta $ from both sides, we get, $\cos 2\theta + 2{\sin ^2}\theta - \cos 2\theta = 1 - \cos 2\theta $ , i.e., $2{\sin ^2}\theta = 1 - \cos 2\theta $ . Now, finally, divide both sides by $2$ , we get, \[{\sin ^2}\theta = \dfrac{{1 - \cos 2\theta }}{2}\] .
Similarly, we can also have $\cos 2\theta = 2{\cos ^2}\theta - 1$ , adding $1$ on both sides, we get, $\cos 2\theta + 1 = 2{\cos ^2}\theta - 1 + 1$ , i.e., $\cos 2\theta + 1 = 2{\cos ^2}\theta $ . Now, dividing, both sides by $2$ , we get, $\dfrac{{\cos 2\theta + 1}}{2} = \dfrac{{2{{\cos }^2}\theta }}{2}$ , which can also be written as ${\cos ^2}\theta = \dfrac{{1 + \cos 2\theta }}{2}$ .
Put these values in \[{\cot ^2}\theta = \dfrac{{{{\cos }^2}\theta }}{{{{\sin }^2}\theta }}\] , we get, \[{\cot ^2}\theta = \dfrac{{\dfrac{{1 + \cos 2\theta }}{2}}}{{\dfrac{{1 - \cos 2\theta }}{2}}}\] , i.e., \[{\cot ^2}\theta = \dfrac{{\left( {1 + \cos 2\theta } \right) \times 2}}{{\left( {1 - \cos 2\theta } \right) \times 2}}\] , now $2$ will be canceled out by $2$ , then we get, \[{\cot ^2}\theta = \dfrac{{1 + \cos 2\theta }}{{1 - \cos 2\theta }}\] .
Hence, the non-exponential trigonometric function of ${\cot ^2}\theta $ is \[\dfrac{{1 + \cos 2\theta }}{{1 - \cos 2\theta }}\] .

Note:
Non- exponential function simply means the resultant function should not have a degree of more than one, i.e., the highest power must be equal to one.
One must have knowledge of all the basic identities associated with the trigonometric functions.
These types of questions are a bit tricky and difficult, so one can do silly mistakes if not done with full concentration.