Evaluate $ 4\cos {20^ \circ } - \sqrt 3 cot{20^ \circ } $

Hint: Cot can also be written as the ratio of cosine to sine. Replace the cot with the ratio of cosine and sine. Use the formulas of $ \sin A\cos B $ and $ \cos A\sin B $ wherever it is necessary. And also remember that sine of negative angle is equal to negative sine angle.
Formulas used:
1. $ 2\cos A\sin B = \sin \left( {A + B} \right) - \sin \left( {A - B} \right) $
2. $ 2\sin A\cos B = \sin \left( {A + B} \right) + \sin \left( {A - B} \right) $
3. $ \sin \left( { - \theta } \right) = - \sin \theta $
4. $ \sin A - \sin B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right) $

Complete step-by-step answer:
We are given to evaluate the expression $ 4\cos {20^ \circ } - \sqrt 3 cot{20^ \circ } $
We have cot function in the given expression; express it in terms of cosine and sine as $ \dfrac{{\cos {{20}^ \circ }}}{{\sin {{20}^ \circ }}} $
Therefore, the expression becomes $ 4\cos {20^ \circ } - \sqrt 3 \left( {\dfrac{{\cos {{20}^ \circ }}}{{\sin {{20}^ \circ }}}} \right) $
We are next taking LCM and multiplying it to cosine
 $ \Rightarrow \dfrac{{4\cos {{20}^ \circ }\sin {{20}^ \circ } - \sqrt 3 \cos {{20}^ \circ }}}{{\sin {{20}^ \circ }}} $
Taking $ \sin {20^ \circ } $ out, we get
 $ \Rightarrow \dfrac{1}{{\sin {{20}^ \circ }}}\left( {4\cos {{20}^ \circ }\sin {{20}^ \circ } - \sqrt 3 \cos {{20}^ \circ }} \right) $
 $ \Rightarrow \dfrac{1}{{\sin {{20}^ \circ }}}\left[ {2\left( {2\cos {{20}^ \circ }\sin {{20}^ \circ }} \right) - \sqrt 3 \cos {{20}^ \circ }} \right] $
As we can see the first term inside the bracket is of the form $ 2\cos A\sin B $ which is equal to $ \sin \left( {A + B} \right) - \sin \left( {A - B} \right) $ , where A and B are equal to $ {20^ \circ } $
Therefore, $ 2\cos {20^ \circ }\sin {20^ \circ } = \sin \left( {{{20}^ \circ } + {{20}^ \circ }} \right) - \sin \left( {{{20}^ \circ } - {{20}^ \circ }} \right) = \sin {40^ \circ } - 0 = \sin {40^ \circ } $
On substituting the obtained value of $ 2\cos {20^ \circ }\sin {20^ \circ } $ , we get
 $ \Rightarrow \dfrac{1}{{\sin {{20}^ \circ }}}\left[ {2\sin {{40}^ \circ } - \sqrt 3 \cos {{20}^ \circ }} \right] $
Taking 2 out common, we get
 $ \Rightarrow \dfrac{2}{{\sin {{20}^ \circ }}}\left[ {\sin {{40}^ \circ } - \dfrac{{\sqrt 3 }}{2}\cos {{20}^ \circ }} \right] $
 $ \dfrac{{\sqrt 3 }}{2} $ can also be written as $ \sin {60^ \circ } $
Therefore, the expression becomes $ \Rightarrow \dfrac{2}{{\sin {{20}^ \circ }}}\left[ {\sin {{40}^ \circ } - \sin {{60}^ \circ }\cos {{20}^ \circ }} \right] $
Now we are sending 2 back inside
 $ \Rightarrow \dfrac{1}{{\sin {{20}^ \circ }}}\left[ {2\sin {{40}^ \circ } - 2\sin {{60}^ \circ }\cos {{20}^ \circ }} \right] $
As we can see the second term inside the bracket is in the form $ 2\sin A\cos B $ which is equal to $ \sin \left( {A + B} \right) + \sin \left( {A - B} \right) $ , where A is $ {60^ \circ } $ and B is $ {20^ \circ } $
Therefore, $ 2\sin {60^ \circ }\cos {20^ \circ } = \sin \left( {{{60}^ \circ } + {{20}^ \circ }} \right) + \sin \left( {{{60}^ \circ } - {{20}^ \circ }} \right) = \sin {80^ \circ } + \sin {40^ \circ } $
On substituting the obtained value of $ 2\sin {60^ \circ }\cos {20^ \circ } $ , we get
 $ \Rightarrow \dfrac{1}{{\sin {{20}^ \circ }}}\left[ {2\sin {{40}^ \circ } - \left( {\sin {{80}^ \circ } + \sin {{40}^ \circ }} \right)} \right] $
 $ \Rightarrow \dfrac{1}{{\sin {{20}^ \circ }}}\left[ {2\sin {{40}^ \circ } - \sin {{80}^ \circ } - \sin {{40}^ \circ }} \right] = \dfrac{1}{{\sin {{20}^ \circ }}}\left[ {\sin {{40}^ \circ } - \sin {{80}^ \circ }} \right] $
The above expression in the bracket is in the form $ \sin A - \sin B $ which is equal to $ 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right) $ , where A is $ {40^ \circ } $ and B is $ {80^ \circ } $
Therefore, $ \sin {40^ \circ } - \sin {80^ \circ } = 2\cos \left( {\dfrac{{{{40}^ \circ } + {{80}^ \circ }}}{2}} \right)\sin \left( {\dfrac{{{{40}^ \circ } - {{80}^ \circ }}}{2}} \right) = 2\cos {60^ \circ }\sin \left( { - {{20}^ \circ }} \right) $ .
We know that $ \sin \left( { - \theta } \right) = - \sin \theta $ , this means $ \sin \left( { - {{20}^ \circ }} \right) = - \sin {20^ \circ } $
Therefore, $ \sin {40^ \circ } - \sin {80^ \circ } = - 2\cos {60^ \circ }\sin {20^ \circ } $
The expression becomes
 $ \dfrac{1}{{\sin {{20}^ \circ }}}\left( { - 2\cos {{60}^ \circ }\sin {{20}^ \circ }} \right) = \dfrac{{ - \sin {{20}^ \circ }}}{{\sin {{20}^ \circ }}}\left( {2\cos {{60}^ \circ }} \right) = - 2 \times \cos {60^ \circ } $
The value of $ \cos {60^ \circ } = \dfrac{1}{2} $
Therefore, $ - 2 \times \cos {60^ \circ } = - 2 \times \dfrac{1}{2} = - 1 $
The value of $ 4\cos {20^ \circ } - \sqrt 3 cot{20^ \circ } $ is -1.
So, the correct answer is “-1”.

Note: Cot is the inverse of tan, tan is the ratio of sine and cosine. Do not confuse that cot is the ratio of sine and cosine; rather it is the ratio of cosine and sine. For these types of questions, one needs to know all the formulas involving sine and cosine. Be careful with the signs of the terms while writing the formulas.