If $\theta $ is the angle between unit vectors $\overrightarrow A $ and $\overrightarrow B $ , then $\left( {\dfrac{{1 - \overrightarrow A .\overrightarrow B }}{{1 + \overrightarrow A .\overrightarrow B }}} \right)$ is equal to
(A) ${\tan ^2}\left( {\dfrac{\theta }{2}} \right)$
(B) ${\sin ^2}\left( {\dfrac{\theta }{2}} \right)$
(C) ${\cot ^2}\left( {\dfrac{\theta }{2}} \right)$
(D) ${\cos ^2}\left( {\dfrac{\theta }{2}} \right)$

Hint: To solve this question, you first need to understand that since both the vectors are unit vectors, their magnitudes will be $1$ . Hence, the dot product of both the vectors will be nothing but the cosine of the angle between the vectors. To further solve the question, you also need to know the properties of trigonometry:
$2{\sin ^2}\left( {\dfrac{\theta }{2}} \right) = 1 - \cos \theta $ and $2{\cos ^2}\left( {\dfrac{\theta }{2}} \right) = 1 + \cos \theta $

Complete step by step solution:
We will proceed with the same approach as mentioned in the hint section of the solution to the question asked to us. Let us first have a look at what is given to us in the question and what does it mean:
Both the vectors $\overrightarrow A $ and $\overrightarrow B $ are unit vectors, hence, we can confidently say that:
$\left| {\overrightarrow A } \right| = \left| {\overrightarrow B } \right| = 1$
The angle between both the given unit vectors is given to be $\theta $
Now, we can define the dot-product of both the vectors as:
$\overrightarrow A .\overrightarrow B = \left| {\overrightarrow A } \right|\left| {\overrightarrow B } \right|\cos \theta $
Substituting $\left| {\overrightarrow A } \right| = \left| {\overrightarrow B } \right| = 1$ in the equation above, we get:
$
  \overrightarrow A .\overrightarrow B = 1 \times 1 \times \cos \theta \\
   \Rightarrow \overrightarrow A .\overrightarrow B = \cos \theta \\
 $
Now, let us have a look at the equation that is given to us in the question itself:
$\dfrac{{1 - \overrightarrow A .\overrightarrow B }}{{1 + \overrightarrow A .\overrightarrow B }}$
If we substitute the value of the dot-products of the vectors as we found out above, $\overrightarrow A .\overrightarrow B = \cos \theta $ , we get:
$\dfrac{{1 - \overrightarrow A .\overrightarrow B }}{{1 + \overrightarrow A .\overrightarrow B }} = \dfrac{{1 - \cos \theta }}{{1 + \cos \theta }}$
Now, we can see that this term can not be further simplified without the use of trigonometric properties of half angles, which are as follows:
$2{\sin ^2}\left( {\dfrac{\theta }{2}} \right) = 1 - \cos \theta $
$2{\cos ^2}\left( {\dfrac{\theta }{2}} \right) = 1 + \cos \theta $
As we can see, we can substitute the above property in the numerator part of the term and the property in the below in the denominator part of the term from the question, this leaves us with:
$\dfrac{{1 - \overrightarrow A .\overrightarrow B }}{{1 + \overrightarrow A .\overrightarrow B }} = \dfrac{{2{{\sin }^2}\left( {\dfrac{\theta }{2}} \right)}}{{2{{\cos }^2}\left( {\dfrac{\theta }{2}} \right)}}$
We can clearly see that the two in both numerator and denominator can be cancelled out, furthermore, we also see that:
$\dfrac{{{{\sin }^2}\left( {\dfrac{\theta }{2}} \right)}}{{{{\cos }^2}\left( {\dfrac{\theta }{2}} \right)}} = {\tan ^2}\left( {\dfrac{\theta }{2}} \right)$
Putting this is the equation, we get:
$\dfrac{{1 - \overrightarrow A .\overrightarrow B }}{{1 + \overrightarrow A .\overrightarrow B }} = {\tan ^2}\left( {\dfrac{\theta }{2}} \right)$

Hence, we can see that the correct option to the question is nothing but the option (A) as it matches with the value that we got after solving the question.

Note: Many students neglect the important information of both the vectors being unit vectors. Thus, they have the magnitude of both the vectors in their expression while solving the question and get stuck in the process. Also, you must remember the important trigonometric properties to help you solve such questions faster and efficiently.