Answer
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Hint: Here, we will use the reciprocal trigonometric function of \[\cot x\]. Then we will write the angle of the reciprocal trigonometric function as the sum of two angles. Then we will find the value of the reciprocal trigonometric function by using the suitable trigonometric formula. Finally, we will take the reciprocal of the obtained value to get the required answer.
Formula used:
\[\tan (a + b) = \dfrac{{\tan a + \tan b}}{{1 - \tan a\tan b}}\]
Complete step by step solution:
We are required to evaluate \[\cot \left( {\dfrac{{4\pi }}{3}} \right)\].
Let us convert this to a different ratio. We know that the reciprocal ratio of \[\cot (x)\] is the ratio \[\tan (x)\].
So, we will first evaluate \[\tan \left( {\dfrac{{4\pi }}{3}} \right)\] and then take the reciprocal of the resultant value to get the value of \[\cot \left( {\dfrac{{4\pi }}{3}} \right)\].
Now, we can express the angle \[\dfrac{{4\pi }}{3}\] as a sum of two known angles, which are \[\pi \] and \[\dfrac{\pi }{3}\]. We can write this as:
\[\dfrac{{4\pi }}{3} = \pi + \dfrac{\pi }{3}\]
So, we will find the value of \[\tan \left( {\pi + \dfrac{\pi }{3}} \right)\]. To do this, let us apply the trigonometric identity \[\tan (a + b) = \dfrac{{\tan a + \tan b}}{{1 - \tan a\tan b}}\]. Here, we will take \[a = \pi \] and \[b = \dfrac{\pi }{3}\].
Substituting these values in the identity, we get
\[\tan \left( {\pi + \dfrac{\pi }{3}} \right) = \dfrac{{\tan \pi + \tan \dfrac{\pi }{3}}}{{1 - \tan \pi \tan \dfrac{\pi }{3}}}\] ……… \[\left( 1 \right)\]
We know that \[\tan \pi = 0\] and \[\tan \dfrac{\pi }{3} = \sqrt 3 \]. Using these values in above equation, we have
\[ \Rightarrow \tan \left( {\pi + \dfrac{\pi }{3}} \right) = \dfrac{{0 + \sqrt 3 }}{{1 - (0 \times \sqrt 3 )}}\]
The numerator becomes equal to \[\sqrt 3 \] and the denominator becomes equal to 1.
Hence, we have the value of \[\tan \left( {\dfrac{{4\pi }}{3}} \right)\] as
\[ \Rightarrow \tan \left( {\dfrac{{4\pi }}{3}} \right) = \dfrac{{\sqrt 3 }}{1} = \sqrt 3 \] ………\[(2)\]
To find the value of \[\cot \left( {\dfrac{{4\pi }}{3}} \right)\], we must find the reciprocal of the value in equation \[(2)\]. Thus, we get
\[\cot \left( {\dfrac{{4\pi }}{3}} \right) = \dfrac{1}{{\sqrt 3 }}\]
Let us rationalize the value in the above equation. To do this we will multiply the numerator and denominator by \[\sqrt 3 \]. This gives us
\[ \Rightarrow \cot \left( {\dfrac{{4\pi }}{3}} \right) = \dfrac{1}{{\sqrt 3 }} \times \dfrac{{\sqrt 3 }}{{\sqrt 3 }} = \dfrac{{\sqrt 3 }}{3}\]
Note:
An alternate way of solving the above problem is to express the ratio \[\cot (x)\] as \[\dfrac{{\cos (x)}}{{\sin (x)}}\].
The value of \[\cos \left( {\dfrac{{4\pi }}{3}} \right) = \cos \left( {\pi + \dfrac{\pi }{3}} \right) = - \dfrac{1}{2}\] and the value of \[\sin \left( {\dfrac{{4\pi }}{3}} \right) = \sin \left( {\pi + \dfrac{\pi }{3}} \right) = - \dfrac{{\sqrt 3 }}{2}\]
Dividing these values, we have
\[\cot \left( {\dfrac{{4\pi }}{3}} \right) = \dfrac{{\left( { - \dfrac{1}{2}} \right)}}{{\left( { - \dfrac{{\sqrt 3 }}{2}} \right)}} = \dfrac{1}{{\sqrt 3 }}\]
Rationalizing the value on the RHS, we get the same value as the above method.
Formula used:
\[\tan (a + b) = \dfrac{{\tan a + \tan b}}{{1 - \tan a\tan b}}\]
Complete step by step solution:
We are required to evaluate \[\cot \left( {\dfrac{{4\pi }}{3}} \right)\].
Let us convert this to a different ratio. We know that the reciprocal ratio of \[\cot (x)\] is the ratio \[\tan (x)\].
So, we will first evaluate \[\tan \left( {\dfrac{{4\pi }}{3}} \right)\] and then take the reciprocal of the resultant value to get the value of \[\cot \left( {\dfrac{{4\pi }}{3}} \right)\].
Now, we can express the angle \[\dfrac{{4\pi }}{3}\] as a sum of two known angles, which are \[\pi \] and \[\dfrac{\pi }{3}\]. We can write this as:
\[\dfrac{{4\pi }}{3} = \pi + \dfrac{\pi }{3}\]
So, we will find the value of \[\tan \left( {\pi + \dfrac{\pi }{3}} \right)\]. To do this, let us apply the trigonometric identity \[\tan (a + b) = \dfrac{{\tan a + \tan b}}{{1 - \tan a\tan b}}\]. Here, we will take \[a = \pi \] and \[b = \dfrac{\pi }{3}\].
Substituting these values in the identity, we get
\[\tan \left( {\pi + \dfrac{\pi }{3}} \right) = \dfrac{{\tan \pi + \tan \dfrac{\pi }{3}}}{{1 - \tan \pi \tan \dfrac{\pi }{3}}}\] ……… \[\left( 1 \right)\]
We know that \[\tan \pi = 0\] and \[\tan \dfrac{\pi }{3} = \sqrt 3 \]. Using these values in above equation, we have
\[ \Rightarrow \tan \left( {\pi + \dfrac{\pi }{3}} \right) = \dfrac{{0 + \sqrt 3 }}{{1 - (0 \times \sqrt 3 )}}\]
The numerator becomes equal to \[\sqrt 3 \] and the denominator becomes equal to 1.
Hence, we have the value of \[\tan \left( {\dfrac{{4\pi }}{3}} \right)\] as
\[ \Rightarrow \tan \left( {\dfrac{{4\pi }}{3}} \right) = \dfrac{{\sqrt 3 }}{1} = \sqrt 3 \] ………\[(2)\]
To find the value of \[\cot \left( {\dfrac{{4\pi }}{3}} \right)\], we must find the reciprocal of the value in equation \[(2)\]. Thus, we get
\[\cot \left( {\dfrac{{4\pi }}{3}} \right) = \dfrac{1}{{\sqrt 3 }}\]
Let us rationalize the value in the above equation. To do this we will multiply the numerator and denominator by \[\sqrt 3 \]. This gives us
\[ \Rightarrow \cot \left( {\dfrac{{4\pi }}{3}} \right) = \dfrac{1}{{\sqrt 3 }} \times \dfrac{{\sqrt 3 }}{{\sqrt 3 }} = \dfrac{{\sqrt 3 }}{3}\]
Note:
An alternate way of solving the above problem is to express the ratio \[\cot (x)\] as \[\dfrac{{\cos (x)}}{{\sin (x)}}\].
The value of \[\cos \left( {\dfrac{{4\pi }}{3}} \right) = \cos \left( {\pi + \dfrac{\pi }{3}} \right) = - \dfrac{1}{2}\] and the value of \[\sin \left( {\dfrac{{4\pi }}{3}} \right) = \sin \left( {\pi + \dfrac{\pi }{3}} \right) = - \dfrac{{\sqrt 3 }}{2}\]
Dividing these values, we have
\[\cot \left( {\dfrac{{4\pi }}{3}} \right) = \dfrac{{\left( { - \dfrac{1}{2}} \right)}}{{\left( { - \dfrac{{\sqrt 3 }}{2}} \right)}} = \dfrac{1}{{\sqrt 3 }}\]
Rationalizing the value on the RHS, we get the same value as the above method.
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