Question

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.$\cot 2A$

Answer 1

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
$= \cot 2A$