Question

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) \[\dfrac{4}{n\left( n+1 \right)\pi }\]

Answer 1

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.