What is the value of $\dfrac{1}{{\cos 80}} - \dfrac{{\sqrt 3 }}{{\sin 80}}$?
A. $\sqrt 2 $
B. $\sqrt 3 $
C. $2$
D. $4$

Hint: We know that $\sin 60 = \dfrac{{\sqrt 3 }}{2}$ and $\cos 60 = \dfrac{1}{2}$so if we take LCM and divide and multiply by 2 in denominator and numerator ,we can get terms like $\sin A\cos B - \cos A\sin B$ where $A = 60$ and $B = 80$.From here we can simplify the fraction using some basic formula of trigonometry to get answer.

Formula Used:
1.$\sin (A - B) = \sin A\cos B - \cos A\sin B$
2.$\sin 2A = 2\sin A\cos A$
3.$\sin (180 - \theta ) = \sin \theta $

Complete step by step solution:
Given -$\dfrac{1}{{\cos 80}} - \dfrac{{\sqrt 3 }}{{\sin 80}}$
Taking LCM and simplifying the numerator and denominator
$\dfrac{1}{{\cos 80}} - \dfrac{{\sqrt 3 }}{{\sin 80}} = \dfrac{{\sin 80 - \sqrt 3 \cos 80}}{{\cos 80\sin 80}}$
Multiplying and dividing by 2
$\dfrac{1}{{\cos 80}} - \dfrac{{\sqrt 3 }}{{\sin 80}} = \dfrac{2}{2}\dfrac{{\sin 80 - \sqrt 3 \cos 80}}{{\cos 80\sin 80}}$
Rearranging the terms of numerator and denominator
$\dfrac{1}{{\cos 80}} - \dfrac{{\sqrt 3 }}{{\sin 80}} = \dfrac{1}{2}\left( {\sin 80 - \sqrt 3 \cos 80} \right)*\dfrac{2}{{\sin 80\cos 80}}$
On simplifying the terms on RHS
RHS=$\left( {\dfrac{1}{2}\sin 80 - \dfrac{{\sqrt 3 }}{2}\cos 80} \right)\dfrac{2}{{\sin 80\cos 80}}$
RHS=$\left( {\cos 60\sin 80 - \sin 60\cos 80} \right)\dfrac{2}{{\sin 80\cos 80}}$
Using the formula $\sin (A - B) = \sin A\cos B - \cos A\sin B$
RHS=$\left( {\sin (80 - 60)} \right)\dfrac{2}{{\sin 80\cos 80}}$
⇒RHS=$\sin 20\dfrac{2}{{\sin 80\cos 80}}$
⇒RHS=$\dfrac{{2\sin 20}}{{\sin 80\cos 80}}$
We know that $\sin 2A = 2\sin A\cos A$
So $\sin A\cos A = \dfrac{{\sin 2A}}{2}$
Hence $\sin 80\cos 80 = \dfrac{{\sin 2 \times 80}}{2}$
$ \Rightarrow \sin 80\cos 80 = \dfrac{{\sin 160}}{2}$
Now RHS =$\sin 20 \times \dfrac{2}{{\dfrac{{\sin 160}}{2}}}$
$4\dfrac{{\sin 20}}{{\sin 180}}$⇒RHS =$4\dfrac{{\sin 20}}{{\sin 160}}$
We know that $\sin (180 - \theta ) = \sin \theta $,put $\theta = 20$in the identity
Then $\sin (180 - 20) = \sin 20$
           $ \Rightarrow \sin 160 = \sin 20$
            So RHS =4
∴ $\dfrac{1}{{\cos 80}} - \dfrac{{\sqrt 3 }}{{\sin 80}} = 4$

Option ‘D’ is correct

Note: In the questions involving complicated trigonometric values which are not known to us ,approach is try to manipulate the question by dividing both side of numerator and denominator by a suitable number or applying trigonometric formulas to convert them into a known value and then solve the problem.