f $ \cos \left( \alpha -\beta \right)=1 $ and $ \cos \left( \alpha +\beta \right)=\dfrac{1}{e} $ , where $ \alpha ,\beta \in \left[ -\pi ,\pi \right] $ . Then the number of ordered pairs of $ \left( \alpha ,\beta \right) $ is:
A. 0
B. 1
C. 2
D. 4

Hint: We can solve this question by forming one a relation between $ \alpha $ and $ \beta $ from one of the given trigonometric equations and using that relation, put the value of one variable in terms of the other variable and put it into the other equation and obtain its different values in the given domain. Here, we will use the equation $ \cos \left( \alpha -\beta \right)=1 $. This way we will get the number of ordered pairs for $ \left( \alpha ,\beta \right) $

Complete step-by-step answer:
Now, it is given that $ \cos \left( \alpha -\beta \right)=1 $
 $ \Rightarrow \alpha -\beta =2n\pi $
Now, since its given in the question that $ \alpha ,\beta \in \left[ -\pi ,\pi \right] $ ,we can form the following relations:
 $ -\pi \le \alpha \le \pi $ .....(1)
 $ \begin{align}
  & -\pi \le \beta \le \pi \\
 & \Rightarrow \pi \ge -\beta \ge -\pi \\
\end{align} $
 $ \Rightarrow -\pi \le -\beta \le \pi $ .....(2)
Now we will add the inequalities (1) and (2) and as a result we will get:
 $ -2\pi \le \alpha -\beta \le 2\pi $
Since we have $ \alpha -\beta =2n\pi $
 $ -2\pi \le 2n\pi \le 2\pi $
 $ \begin{align}
  & \Rightarrow -1\le n\le 1 \\
 & \Rightarrow n\in \{-1,0,1\} \\
\end{align} $
Therefore, we get $ \alpha -\beta \in \{-2\pi ,0,2\pi \} $
Now, we can write $ \alpha $ in terms of $ \beta $ as:
\[\Rightarrow \alpha \in \{\beta -2\pi ,\beta ,\beta +2\pi \}\]
  Now we can calculate $ \alpha +\beta $ as:
\[\Rightarrow \alpha +\beta \in \{2\beta -2\pi ,2\beta ,2\beta +2\pi \}\]
Now, $ \cos (\alpha +\beta ) $ can be calculated as:
  $ \Rightarrow \cos \left( \alpha +\beta \right)=\{\cos \left( 2\beta -2\pi \right),\cos \left( 2\beta \right),\cos \left( 2\beta +2\pi \right)\} $
Since $ \cos \left( 2\beta -2\pi \right)=\cos 2\beta =\cos \left( 2\beta +2\pi \right) $
Therefore, $ \cos \left( \alpha +\beta \right)=\cos 2\beta $ and $ \alpha =\beta $
Now, it is given that $ \cos (\alpha +\beta )=\dfrac{1}{e} $
Thus, $ \cos 2\beta =\dfrac{1}{e} $ which is less than 1
Since, \[\beta \in \left[ -\pi ,\pi \right]\Rightarrow 2\beta \in \left[ -2\pi ,2\pi \right]\] and also, $ \dfrac{1}{e}<1 $
Therefore, there are 4 values of $ \beta $ between \[\left[ -2\pi ,2\pi \right]\] for which it satisfies the given two trigonometric equations as ‘cos’ function has values belonging to $ \left[ -1,1 \right] $ and each value is given by any two angles(in radians) belonging to \[\left[ -0,2\pi \right]\] and since \[2\beta \in \left[ -2\pi ,2\pi \right]\],there will be a total of ‘4’ values for \[\beta \] satisfying the given equations.
Now, we have already established that $ \alpha =\beta $
Therefore, for every value of \[\beta \] ,there is one equal value of $ \alpha $
Thus, there are a total of 4 ordered pairs for $ \left( \alpha ,\beta \right) $ satisfying the given conditions.
So, the correct answer is “Option D”.

Note: We can also establish a relation between $ \alpha $ and $ \beta $ by using the other given trigonometric equation but it will result in the use of inverse cosine function which will make the calculation really complicated and long which result in a lot of wastage of time. Even after that long and tedious calculation, we may get an answer but it has a very high possibility of being wrong as that complicated calculation will increase the scope for committing mistakes. Hence the equation used above should be used in establishing the relation between $ \alpha $ and $ \beta $ .