Solve \[\left[ {\dfrac{1}{{\tan 3A - \tan A}}} \right] - \left[ {\dfrac{1}{{\cot 3A - \cot A}}} \right] = \]
A.\[\tan A\]
B.\[\tan 2A\]
C.\[\cot A\]
D.\[\cot 2A\]

Hint: For question related to proof of \[\tan \] and \[\cot \] we have always have to convert \[\cot \] and \[\tan \] into \[\tan = \dfrac{{\sin }}{{\cos }}\] and for \[\cot = \dfrac{{\cos }}{{\sin }}\] , which will simplify the question briefly . Do not use any identity for \[\tan 3A\] and \[\cot 3A\] which would make the question more complex . These questions are short and tricky .

Complete step-by-step answer:
 Given : \[\left[ {\dfrac{1}{{\tan 3A - \tan A}}} \right] - \left[ {\dfrac{1}{{\cot 3A - \cot A}}} \right]\] ,
putting \[\tan 3A = \dfrac{{\sin 3A}}{{\cos 3A}}\] and \[\cot 3A = \dfrac{{\cos 3A}}{{\sin 3A}}\] we get ,
\[ = \left[ {\dfrac{1}{{\dfrac{{\sin 3A}}{{\cos 3A}} - \dfrac{{\sin A}}{{\cos A}}}}} \right] - \left[ {\dfrac{1}{{\dfrac{{\cos 3A}}{{\sin 3A}} - \dfrac{{\cos A}}{{\sin A}}}}} \right]\] , on solving we get
\[ = \left[ {\dfrac{1}{{\dfrac{{\sin 3A\cos A - \cos 3A\sin A}}{{\cos 3A{\text{ }}\cos A}}}}} \right] - \left[ {\dfrac{1}{{\dfrac{{\cos 3A\sin A - \sin 3A\cos A}}{{\sin 3A\sin A}}}}} \right]\]
On further simplifying, we get
\[ = \left[ {\dfrac{{\cos 3A\cos A}}{{\sin 3A\cos A - \cos 3A\sin A}}} \right] - \left[ {\dfrac{{\sin 3A\sin A}}{{\cos 3A\sin A - \sin 3A\cos A}}} \right]\] , on rearranging we get ,
\[ = \left[ {\dfrac{{\cos 3A\cos A}}{{\sin 3A\cos A - \cos 3A\sin A}}} \right] + \left[ {\dfrac{{\sin 3A\sin A}}{{\sin 3A\cos A - \cos 3A\sin A}}} \right]\]
Now , using the identity of \[\sin (A + B) = \sin A\cos B + \cos A\sin B\] in denominator we get ,
\[ = \left[ {\dfrac{{\cos 3A\cos A}}{{\sin (3A - A)}}} \right] + \left[ {\dfrac{{\sin 3A\sin A}}{{\sin (3A - A)}}} \right]\] , on solving we get
\[ = \left[ {\dfrac{{\cos 3A\cos A}}{{\sin 2A}}} \right] + \left[ {\dfrac{{\sin 3A\sin A}}{{\sin 2A}}} \right]\]
On solving further we get
\[ = \left[ {\dfrac{{\cos 3A\cos A + \sin 3A\sin A}}{{\sin 2A}}} \right]\] ,
 Now using the identity of \[\cos (A - B) = \cos A\cos B + \sin A\sin B\] , we get
\[ = \left[ {\dfrac{{\cos (3A - A)}}{{\sin 2A}}} \right]\]
\[ = \left[ {\dfrac{{\cos 2A}}{{\sin 2A}}} \right]\]
Hence, we can write the final answer as
\[ = \cot 2A\] .
Therefore , option ( 4 ) is the correct answer for the given question .
So, the correct answer is “Option 4”.

Note: Alternate Method :
\[ = \left[ {\dfrac{1}{{\tan 3A - \tan A}}} \right] - \left[ {\dfrac{1}{{\cot 3A - \cot A}}} \right]\]
We know that \[\cot A = \dfrac{1}{{\tan A}}\] , applying this to above equation we get
\[ = \left[ {\dfrac{1}{{\tan 3A - \tan A}}} \right] - \left[ {\dfrac{1}{{\dfrac{1}{{\tan 3A}} - \dfrac{1}{{\tan A}}}}} \right]\]
On solving we get
\[ = \left[ {\dfrac{1}{{\tan 3A - \tan A}}} \right] - \left[ {\dfrac{1}{{\dfrac{1}{{\tan 3A}} - \dfrac{1}{{\tan A}}}}} \right]\] , on solving we get
\[ = \left[ {\dfrac{1}{{\tan 3A - \tan A}}} \right] - \left[ {\dfrac{1}{{\dfrac{{\tan A - \tan 3A}}{{\tan 3A\tan A}}}}} \right]\]
On simplifying we get
\[ = \left[ {\dfrac{1}{{\tan 3A - \tan A}}} \right] - \left[ {\dfrac{{\tan 3A\tan A}}{{\tan A - \tan 3A}}} \right]\] , on rearranging we get
\[ = \left[ {\dfrac{1}{{\tan 3A - \tan A}}} \right] + \left[ {\dfrac{{\tan 3A\tan A}}{{\tan 3A - \tan A}}} \right]\]
On further solving we get
\[ = \left[ {\dfrac{{1 + \tan 3A\tan A}}{{\tan 3A - \tan A}}} \right]\] , rewriting the expression we get
\[ = \left[ {\dfrac{1}{{\dfrac{{\tan 3A - \tan A}}{{1 + \tan 3A\tan A}}}}} \right]\]
Now using the identity of \[\tan (A + B) = \dfrac{{\tan A + \tan B}}{{1 + \tan A\tan B}}\]
\[ = \left[ {\dfrac{1}{{\tan (3A - A)}}} \right]\] , on solving we get
\[ = \left[ {\dfrac{1}{{\tan 2A}}} \right]\]
Hence, we can write the final answer as
\[ = \cot 2A\]