
What is the exact value of \[\cot\ 210^{o}\] ?
Answer
476.7k+ views
Hint: In this question, we need to find the value of \[\cot\ 210^{o}\] . In this type of question we have to use the concepts of trigonometry. We know that the cotangent is the ratio of the cosine and sine functions so we can write this cotangent expression as the ratio of cosine and sine. Then we can rewrite the angle \[210^{o}\] in the form of \[(180^{o} + \theta)\] . Then by using trigonometric identities , we can simplify the expression. Using the value of \[\sin(30^{o})\] and \[\cos(30^{o})\] , we can find the value of cot \[210^{o}\] .
Identities used :
1. \[\sin(180^{o} + \theta)\ = - \sin\theta\]
2. \[\cos\left( 180^{o} + \theta \right) = - \cos\theta\]
Trigonometry table :
Complete step-by-step answer:
Given, \[\cot\ 210^{o}\]
Here we need to find the exact value of \[\cot\ 210^{o}\] .
We know that the cotangent function is the ratio of the cosine and sine functions.
\[\cot\theta = \dfrac{\cos\theta}{\sin\theta}\]
Here angle \[\theta\] is \[210^{o}\] .
\[\cot\ 210^{o} = \dfrac{\cos\left( 210^{o} \right)}{\sin\left( 210^{o} \right)}\]
We can rewrite \[210^{o}\] as \[(180^{o}+ 30^{o})\] ,
We get,
\[\Rightarrow \ \cot\ 210^{o} = \dfrac{\cos\left( 180^{o} + 30^{o} \right)}{\sin\left( 180^{o} + 30^{o} \right)}\]
By using the identities , \[\sin(180^{o} + \theta)\ = - \sin\theta\] and \[\cos(180^{o} + \theta) = - \cos\theta\]
We can write,
\[\Rightarrow \ \cot\ 210^{o} = \dfrac{- \cos\left( 30^{o} \right)}{- \sin\left( 30^{o} \right)}\]
From the trigonometric table, the value of \[\sin(30^{o})\] is \[\dfrac{1}{2}\] and the value of \[\cos(30^{o})\] is \[\dfrac{\sqrt{3}}{2}\] .
\[\Rightarrow \ \cot\ 210^{o} = \dfrac{- \dfrac{\sqrt{3}}{2}}{- \dfrac{{1}}{2}}\]
On multiplying the numerator by the reciprocal of the denominator,
We get,
\[\Rightarrow \ \cot\ 210^{o} = \left( - \dfrac{\sqrt{3}}{2} \right) \times \left( - \dfrac{2}{{1}} \right)\]
On simplifying,
We get,
\[\cot\ 210^{o} = \dfrac{ - \sqrt{3}}{-1} \]
On further simplifying,
We get,
\[\cot\ 210^{o} = \sqrt{3}\]
Thus the value of \[\cot\ 210^{o}\] is equal to \[\sqrt{3}\] .
Final answer :
The exact value of \[\cot\ 210^{o}\] is equal to \[\sqrt{3}\] .
Note: The concept used in this problem is use of trigonometric identities and ratios. The given angle \[210^{o}\] will be in the third quadrant and in the third quadrant tangent and cotangent functions will be positive. Also we have to be careful while taking the value of \[\sin(30^{o})\] and \[\cos(30^{o})\] . If the value of \[\sin(30^{o})\] and \[\cos(30^{o})\] is not known then it is hard to solve this question. If it is not known then it is hard to solve this question.
Identities used :
1. \[\sin(180^{o} + \theta)\ = - \sin\theta\]
2. \[\cos\left( 180^{o} + \theta \right) = - \cos\theta\]
Trigonometry table :
| Angles | \[0^{o}\] | \[30^{o}\] | \[45^{o}\] | \[60^{o}\] | \[90^{o}\] |
| Sine | \[0\] | \[\dfrac{1}{2}\] | \[\dfrac{1}{\sqrt{2}}\] | \[\dfrac{\sqrt{3}}{2}\] | \[1\] |
| Cosine | \[1\] | \[\dfrac{\sqrt{3}}{2}\] | \[\dfrac{1}{\sqrt{2}}\] | \[\dfrac{1}{2}\] | \[0\] |
Complete step-by-step answer:
Given, \[\cot\ 210^{o}\]
Here we need to find the exact value of \[\cot\ 210^{o}\] .
We know that the cotangent function is the ratio of the cosine and sine functions.
\[\cot\theta = \dfrac{\cos\theta}{\sin\theta}\]
Here angle \[\theta\] is \[210^{o}\] .
\[\cot\ 210^{o} = \dfrac{\cos\left( 210^{o} \right)}{\sin\left( 210^{o} \right)}\]
We can rewrite \[210^{o}\] as \[(180^{o}+ 30^{o})\] ,
We get,
\[\Rightarrow \ \cot\ 210^{o} = \dfrac{\cos\left( 180^{o} + 30^{o} \right)}{\sin\left( 180^{o} + 30^{o} \right)}\]
By using the identities , \[\sin(180^{o} + \theta)\ = - \sin\theta\] and \[\cos(180^{o} + \theta) = - \cos\theta\]
We can write,
\[\Rightarrow \ \cot\ 210^{o} = \dfrac{- \cos\left( 30^{o} \right)}{- \sin\left( 30^{o} \right)}\]
From the trigonometric table, the value of \[\sin(30^{o})\] is \[\dfrac{1}{2}\] and the value of \[\cos(30^{o})\] is \[\dfrac{\sqrt{3}}{2}\] .
\[\Rightarrow \ \cot\ 210^{o} = \dfrac{- \dfrac{\sqrt{3}}{2}}{- \dfrac{{1}}{2}}\]
On multiplying the numerator by the reciprocal of the denominator,
We get,
\[\Rightarrow \ \cot\ 210^{o} = \left( - \dfrac{\sqrt{3}}{2} \right) \times \left( - \dfrac{2}{{1}} \right)\]
On simplifying,
We get,
\[\cot\ 210^{o} = \dfrac{ - \sqrt{3}}{-1} \]
On further simplifying,
We get,
\[\cot\ 210^{o} = \sqrt{3}\]
Thus the value of \[\cot\ 210^{o}\] is equal to \[\sqrt{3}\] .
Final answer :
The exact value of \[\cot\ 210^{o}\] is equal to \[\sqrt{3}\] .
Note: The concept used in this problem is use of trigonometric identities and ratios. The given angle \[210^{o}\] will be in the third quadrant and in the third quadrant tangent and cotangent functions will be positive. Also we have to be careful while taking the value of \[\sin(30^{o})\] and \[\cos(30^{o})\] . If the value of \[\sin(30^{o})\] and \[\cos(30^{o})\] is not known then it is hard to solve this question. If it is not known then it is hard to solve this question.
Recently Updated Pages
Master Class 11 Chemistry: Engaging Questions & Answers for Success

Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

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

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

