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Question: If tan (cot x) = cot (tan x), then sin 2x is equal to (a) \[\dfrac{2}{\left( 2n+1 \right)\pi }\] ...

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

Explanation

Solution

First of all, consider the given equation and use cot⁡θ=tan⁡(π2−θ)\cot \theta =\tan \left( \dfrac{\pi }{2}-\theta \right) to convert cot (tan x) into tan⁡(π2−tan⁡x)\tan \left( \dfrac{\pi }{2}-\tan x \right). Now, compare LHS and RHS and get tan⁡x+cot⁡x=nπ+π2\tan x+\cot x=n\pi +\dfrac{\pi }{2}. Now, use tan⁡θ=1cot⁡θ=sin⁡θcos⁡θ\tan \theta =\dfrac{1}{\cot \theta }=\dfrac{\sin \theta }{\cos \theta } and simplify it to get the value of sin⁡2x=2sin⁡xcos⁡x\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⁡(cot⁡x)=cot⁡(tan⁡x)\tan \left( \cot x \right)=\cot \left( \tan x \right)
We know that cot⁡θ=tan⁡(π2−θ)\cot \theta =\tan \left( \dfrac{\pi }{2}-\theta \right).
Now, on RHS we can write cot ( tanx ) as tan⁡(π2−tan⁡x)\tan \left( \dfrac{\pi }{2}-\tan x \right)
tan⁡(cot⁡x)=tan⁡(π2−tan⁡x)\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=π2−tan⁡x\cot x=\dfrac{\pi }{2}-\tan x
tan⁡x+cot⁡x=π2\tan x+\cot x=\dfrac{\pi }{2}
In general form, we can write the above equation as,
tan⁡x+cot⁡x=nπ+π2\tan x+\cot x=n\pi +\dfrac{\pi }{2}
We know that tan⁡θ=sin⁡θcos⁡θ and cot⁡θ=cos⁡θsin⁡θ\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,
sin⁡xcos⁡x+cos⁡xsin⁡x=nπ+π2\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,
sin⁡2x+cos⁡2xcos⁡xsin⁡x=nπ+π2\dfrac{{{\sin }^{2}}x+{{\cos }^{2}}x}{\cos x\sin x}=n\pi +\dfrac{\pi }{2}
We know that sin⁡2θ+cos⁡2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1. By using this, in the above equation, we get,
1cos⁡xsin⁡x=nπ+π2\dfrac{1}{\cos x\sin x}=n\pi +\dfrac{\pi }{2}
1cos⁡xsin⁡x=π(2n+12)\dfrac{1}{\cos x\sin x}=\pi \left( \dfrac{2n+1}{2} \right)
By cross multiplying the above equation, we get,
2π(2n+1)=sin⁡xcos⁡x\dfrac{2}{\pi \left( 2n+1 \right)}=\sin x\cos x
By multiplying 2 on both the sides of the above equation, we get,
4π(2n+1)=2sin⁡xcos⁡x\dfrac{4}{\pi \left( 2n+1 \right)}=2\sin x\cos x
We know that 2sin⁡θcos⁡θ=sin⁡2θ2\sin \theta \cos \theta =\sin 2\theta . By using this in the above equation, we get,
4π(2n+1)=sin⁡2x\dfrac{4}{\pi \left( 2n+1 \right)}=\sin 2x
So, we have got the values of sin 2x as 4π(2n+1)\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.