
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
474.6k+ views
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.
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.
Recently Updated Pages
Master Class 11 Accountancy: Engaging Questions & Answers for Success

Express the following as a fraction and simplify a class 7 maths CBSE

The length and width of a rectangle are in ratio of class 7 maths CBSE

The ratio of the income to the expenditure of a family class 7 maths CBSE

How do you write 025 million in scientific notatio class 7 maths CBSE

How do you convert 295 meters per second to kilometers class 7 maths CBSE

Trending doubts
10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

State and prove Bernoullis theorem class 11 physics CBSE

What organs are located on the left side of your body class 11 biology CBSE

Write down 5 differences between Ntype and Ptype s class 11 physics CBSE
