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**Hint:**First of all, consider the given equation and use \[\cot \theta =\tan \left( \dfrac{\pi }{2}-\theta \right)\] to convert cot (tan x) into \[\tan \left( \dfrac{\pi }{2}-\tan x \right)\]. Now, compare LHS and RHS and get \[\tan x+\cot x=n\pi +\dfrac{\pi }{2}\]. Now, use \[\tan \theta =\dfrac{1}{\cot \theta }=\dfrac{\sin \theta }{\cos \theta }\] and simplify it to get the value of \[\sin 2x=2\sin x\cos x\].

**Complete step by step answer:**

We are given that tan (cot x) = cot (tan x). Then, we have to find the value of sin 2x. Let us consider the equation given in the question.

\[\tan \left( \cot x \right)=\cot \left( \tan x \right)\]

We know that \[\cot \theta =\tan \left( \dfrac{\pi }{2}-\theta \right)\].

Now, on RHS we can write cot ( tanx ) as \[\tan \left( \dfrac{\pi }{2}-\tan x \right)\]

\[\tan \left( \cot x \right)=\tan \left( \dfrac{\pi }{2}-\tan x \right)\]

By comparing RHS and LHS of the above equation, we get,

\[\cot x=\dfrac{\pi }{2}-\tan x\]

\[\tan x+\cot x=\dfrac{\pi }{2}\]

In general form, we can write the above equation as,

\[\tan x+\cot x=n\pi +\dfrac{\pi }{2}\]

We know that \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\text{ and }\cot \theta =\dfrac{\cos \theta }{\sin \theta }\]. By using this in the above equation, we get,

\[\dfrac{\sin x}{\cos x}+\dfrac{\cos x}{\sin x}=n\pi +\dfrac{\pi }{2}\]

By taking sin x cos x as LCM and simplifying the above equation, we get,

\[\dfrac{{{\sin }^{2}}x+{{\cos }^{2}}x}{\cos x\sin x}=n\pi +\dfrac{\pi }{2}\]

We know that \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]. By using this, in the above equation, we get,

\[\dfrac{1}{\cos x\sin x}=n\pi +\dfrac{\pi }{2}\]

\[\dfrac{1}{\cos x\sin x}=\pi \left( \dfrac{2n+1}{2} \right)\]

By cross multiplying the above equation, we get,

\[\dfrac{2}{\pi \left( 2n+1 \right)}=\sin x\cos x\]

By multiplying 2 on both the sides of the above equation, we get,

\[\dfrac{4}{\pi \left( 2n+1 \right)}=2\sin x\cos x\]

We know that \[2\sin \theta \cos \theta =\sin 2\theta \]. By using this in the above equation, we get,

\[\dfrac{4}{\pi \left( 2n+1 \right)}=\sin 2x\]

So, we have got the values of sin 2x as \[\dfrac{4}{\pi \left( 2n+1 \right)}\]

**So, the correct answer is “Option B”.**

**Note:**In this question, students can cross-check their answers by substituting the same value of x and from that finding the value of x. If that value of x satisfies the given equation, then our answer is correct. Also, in this question, some students try to find the value of sin x and cos x individually and then use them to find the value of sin 2x which is not required as we can get sin x cos x easily without calculating each of them individually like in the above solution.

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