Question

How to complete this identity $\dfrac{{\cos \left( {\alpha - \beta } \right)}}{{\cos \alpha \cos \beta }}$?
A. $\tan \alpha \tan \beta + \cot \beta $
B. $1 + \tan \alpha \tan \beta $
C. \[1 + \cot \alpha \tan \beta \]
D. \[1 + \cot \alpha \cot \beta \]

Answer 1

Hint: This problem deals with solving the given equation with trigonometric identities and compound sum angles of trigonometric functions. A compound angle formula or addition formula is a trigonometric identity which expresses a trigonometric function of $\left( {A + B} \right)$ or $\left( {A - B} \right)$in terms of trigonometric functions of $A$ and $B$. The used formula here is:
$ \Rightarrow \cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B$

Complete step-by-step answer:
Given an expression of trigonometric expression functions.
The given expression is $\dfrac{{\cos \left( {\alpha - \beta } \right)}}{{\cos \alpha \cos \beta }}$, consider this as given below:
$ \Rightarrow \dfrac{{\cos \left( {\alpha - \beta } \right)}}{{\cos \alpha \cos \beta }}$
We know the compound angle formula of cosine, hence applying it to the numerator of the given expression, as shown below:
$ \Rightarrow \cos \left( {\alpha - \beta } \right) = \cos \alpha \cos \beta + \sin \alpha \sin \beta $
The given denominator of the expression is given below:
$ \Rightarrow \cos \alpha \cos \beta $
Now substitution the obtained simplified expression of the numerator of the given expression, as shown below:
\[ \Rightarrow \dfrac{{\cos \left( {\alpha - \beta } \right)}}{{\cos \alpha \cos \beta }} = \dfrac{{\cos \alpha \cos \beta + \sin \alpha \sin \beta }}{{\cos \alpha \cos \beta }}\]
Now split the fraction into separate fractions on right hand side of the above equation, as shown below:
\[ \Rightarrow \dfrac{{\cos \left( {\alpha - \beta } \right)}}{{\cos \alpha \cos \beta }} = \dfrac{{\cos \alpha \cos \beta }}{{\cos \alpha \cos \beta }} + \dfrac{{\sin \alpha \sin \beta }}{{\cos \alpha \cos \beta }}\]
Now after splitting the fractions, the first term becomes 1, as the numerator and the denominator are equal.
Now the second term is split in such a way that it can be converted to another trigonometric function:
\[ \Rightarrow \dfrac{{\cos \left( {\alpha - \beta } \right)}}{{\cos \alpha \cos \beta }} = 1 + \left( {\dfrac{{\sin \alpha }}{{\cos \alpha }}} \right)\left( {\dfrac{{\sin \beta }}{{\cos \beta }}} \right)\]
We know that $\dfrac{{\sin A}}{{\cos A}} = \tan A$, applying this identity below:
Here replacing the expression \[\dfrac{{\sin \alpha }}{{\cos \alpha }}\] with $\tan \alpha $, and replacing the expression \[\dfrac{{\sin \beta }}{{\cos \beta }}\] with $\tan \beta $,as shown below:
\[ \Rightarrow \dfrac{{\cos \left( {\alpha - \beta } \right)}}{{\cos \alpha \cos \beta }} = 1 + \tan \alpha \tan \beta \]

Final Answer: The expression is equal to, $\dfrac{{\cos \left( {\alpha - \beta } \right)}}{{\cos \alpha \cos \beta }} = 1 + \tan \alpha \tan \beta $

Note:
Please note that the formula of cosine compound angles formula is used to solve this problem, but there are a few other trigonometric compound angle formulas of sine, cosine and tangent, which are shown below:
$ \Rightarrow \sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B$
$ \Rightarrow \sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B$
$ \Rightarrow \cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B$
$ \Rightarrow \cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B$
$ \Rightarrow \tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$
$ \Rightarrow \tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}$