How to complete the identity $\dfrac{{\cos (\alpha - \beta )}}{{\cos \alpha \cos \beta }} =$ ?
A. $\tan \alpha \tan \beta + \cot \beta$
B. $1 + \tan \alpha \tan \beta$
C. $1 + \cot \alpha \tan \beta$
D. $1 + \cot \alpha \cot \beta$

Hint: To find the complete identity of $\dfrac{{\cos (\alpha - \beta )}}{{\cos \alpha \cos \beta }}$ at first, we will use the formula $\cos \left( {a - b} \right) = \cos a\cos b + \sin a\sin b$ where $a$ and $b$ are arbitrary angle in the numerator. Then we will divide the numerator with the denominator and finally get the answer.

Formula Used:
$\cos \left( {a - b} \right) = \cos a\cos b + \sin a\sin b$ where $a$ and $b$ are arbitrary angles.
$\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ .

Complete step by step answer:
$\dfrac{{\cos (\alpha - \beta )}}{{\cos \alpha \cos \beta }}$
We know that $\cos \left( {a - b} \right) = \cos a\cos b + \sin a\sin b$ where $a$ and $b$ are arbitrary angles. We will apply this formula in the numerator and get;
$\Rightarrow \dfrac{{\cos (\alpha - \beta )}}{{\cos \alpha \cos \beta }} = \dfrac{{\cos \alpha \cos \beta + \sin \alpha \sin \beta }}{{\cos \alpha \cos \beta }}$
Dividing the numerator with the denominator we get;
$\Rightarrow \dfrac{{\cos (\alpha - \beta )}}{{\cos \alpha \cos \beta }} = 1 + \dfrac{{\sin \alpha \sin \beta }}{{\cos \alpha \cos \beta }}$
We know that $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ . Using this in the above equation we get;
$\Rightarrow \dfrac{{\cos (\alpha - \beta )}}{{\cos \alpha \cos \beta }} = 1 + \tan \alpha \tan \beta$
The correct answer is $1 + \tan \alpha \tan \beta$ .

So, the correct answer is Option B.

Note: For the trigonometric derivation, we will try to use the basic formulas in trigonometry to simplify any identity. Here we used the formula $\cos \left( {a - b} \right) = \cos a\cos b + \sin a\sin b$ where $a$ and $b$ are arbitrary angles. After using this formula the identity is easily simplified.
Similar examples:
How to complete the identity $\dfrac{{\cos (\alpha + \beta )}}{{\cos \alpha \cos \beta }} =$ ?

We have given;
$\dfrac{{\cos (\alpha + \beta )}}{{\cos \alpha \cos \beta }}$
We know that $\cos \left( {a + b} \right) = \cos a\cos b - \sin a\sin b$ where $a$ and $b$ are arbitrary angles. We will apply this formula in the numerator and get;
$\Rightarrow \dfrac{{\cos (\alpha + \beta )}}{{\cos \alpha \cos \beta }} = \dfrac{{\cos \alpha \cos \beta - \sin \alpha \sin \beta }}{{\cos \alpha \cos \beta }}$
Dividing the numerator with the denominator we get;
$\Rightarrow \dfrac{{\cos (\alpha + \beta )}}{{\cos \alpha \cos \beta }} = 1 - \dfrac{{\sin \alpha \sin \beta }}{{\cos \alpha \cos \beta }}$
We know that $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ . Using this in the above equation we get;
$\Rightarrow \dfrac{{\cos (\alpha + \beta )}}{{\cos \alpha \cos \beta }} = 1 - \tan \alpha \tan \beta$
How to complete the identity $\dfrac{{\sin (\alpha + \beta )}}{{\sin \alpha \cos \beta }} =$ ?

We have given;
$\dfrac{{\sin (\alpha + \beta )}}{{\sin \alpha \cos \beta }}$
We know that $\sin \left( {a + b} \right) = \sin a\cos b + \cos a\sin b$ where $a$ and $b$ are arbitrary angles. We will apply this formula in the numerator and get;
$\Rightarrow \dfrac{{\sin (\alpha + \beta )}}{{\sin \alpha \cos \beta }} = \dfrac{{\sin \alpha \cos \beta + \cos \alpha \sin \beta }}{{\sin \alpha \cos \beta }}$
Dividing the numerator with the denominator we get;
$\Rightarrow \dfrac{{\sin (\alpha + \beta )}}{{\sin \alpha \cos \beta }} = 1 + \dfrac{{\cos \alpha \sin \beta }}{{\sin \alpha \cos \beta }}$
We know that $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ and $\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$ . Using this in the above equation we get;
$\Rightarrow \dfrac{{\sin (\alpha + \beta )}}{{\sin \alpha \cos \beta }} = 1 + \cot \alpha \tan \beta$
How to complete the identity $\dfrac{{\sin (\alpha - \beta )}}{{\sin \alpha \cos \beta }} =$ ?

We have given;
$\dfrac{{\sin (\alpha - \beta )}}{{\sin \alpha \cos \beta }}$
We know that $\sin \left( {a - b} \right) = \sin a\cos b - \cos a\sin b$ where $a$ and $b$ are arbitrary angles. We will apply this formula in the numerator and get;
$\Rightarrow \dfrac{{\sin (\alpha - \beta )}}{{\sin \alpha \cos \beta }} = \dfrac{{\sin \alpha \cos \beta - \cos \alpha \sin \beta }}{{\sin \alpha \cos \beta }}$
Dividing the numerator with the denominator we get;
$\Rightarrow \dfrac{{\sin (\alpha - \beta )}}{{\sin \alpha \cos \beta }} = 1 - \dfrac{{\cos \alpha \sin \beta }}{{\sin \alpha \cos \beta }}$
We know that $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ and $\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$ . Using this in the above equation we get;
$\Rightarrow \dfrac{{\sin (\alpha - \beta )}}{{\sin \alpha \cos \beta }} = 1 - \cot \alpha \tan \beta$ .