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# If tan (cot x) = cot (tan x), then sin 2x is equal to (a) $\dfrac{2}{\left( 2n+1 \right)\pi }$(b) $\dfrac{4}{\left( 2n+1 \right)\pi }$(c) $\dfrac{2}{n\left( n+1 \right)\pi }$(d) $\dfrac{4}{n\left( n+1 \right)\pi }$

Last updated date: 20th Jun 2024
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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 $\sin 2x=2\sin x\cos x$.

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.