Question

How do I find the value of $\cot 5\pi /6$?

Answer 1

Hint: The cot function (also known as cotangent function) can be defined as the ratio of the length of the base to that of the perpendicular in a right-angled triangle. The formulae for cot function is given as $\cot \theta = \dfrac{{base}}{{perpendicular}}$ where $\theta $ is the angle subtended by the base and the hypotenuse of the respective right-angled triangle.
Also, here we will be using the concept of reference angle. It is the smallest angle that the terminal side of a given angle makes with the $x$-axis. It is always positive and less than ${90^{\circ}}$. Moreover, if we want to find the reference angle for any angle, we will have to first make it lie between ${0^{\circ}}$ and ${360^{\circ}}$ by keeping ${360^{\circ}}$ subtract from it (for negative angles we will add ${360^{\circ}}$). Depending on the quadrant where the angle lies, the reference angle can be valued differently. If the angle is $\theta $ then it will be $\theta $ only if it lies in the first quadrant, ${180^{\circ}} - \theta $ in the second quadrant, ${180^{\circ}} - \theta $ in the third quadrant and ${180^{\circ}} - \theta $ in the fourth quadrant.

Complete step by step solution:
Given the question of which the value is to be found out is $\cot \dfrac{{5\pi }}{6}$, where $\dfrac{{5\pi }}{6}$ is the argument of $\cot $.
We know that $\pi $radians$ = {180^{\circ}}$. On multiplying both sides with $5$, we will get
$ \Rightarrow 5\pi $ radians$ = 5 \times {180^{\circ}}$
On further simplifying, we get
$ \Rightarrow 5\pi $ radians$ = {900^{\circ}}$
Now on dividing both sides by $6$, we will get
$ \Rightarrow \dfrac{{5\pi }}{6}$ radians$ = \dfrac{{{{900}^{\circ}}}}{6}$
On further simplifying, we get
$ \Rightarrow \dfrac{{5\pi }}{6}$ radians$ = {150^{\circ}}$
Hence on converting $\dfrac{{5\pi }}{6}$ to degrees, we got ${150^{\circ}}$. So now we have to find the value of $\cot {150^{\circ}}$. Since ${150^{\circ}} > {90^{\circ}}$, hence we will find the reference angle for ${150^{\circ}}$. As ${150^{\circ}}$ lies in the second quadrant, therefore the reference angle for ${150^{\circ}}$ becomes ${180^{\circ}} - {150^{\circ}} = {30^{\circ}}$.
So now we will be finding the value of $\cot {30^{\circ}}$, such that
$ \Rightarrow \cot {30^{\circ}} = \dfrac{1}{{\tan {{30}^{\circ}}}}$
It must be known that $\tan {30^{\circ}} = \dfrac{1}{{\sqrt 3 }}$. So substituting it below, we get
$ \Rightarrow \cot {30^{\circ}} = \dfrac{1}{{\dfrac{1}{{\sqrt 3 }}}}$
$ \Rightarrow \cot {30^{\circ}} = \sqrt 3 $

Hence, we get that $\cot \dfrac{{5\pi }}{6}$ has the value of $\sqrt 3 $.

Note:
We must know that the tangent of a given angle is equal to the cotangent of its complementary angle. Also, two angles are known to be complementary angles when their sum is equal to ${90^{\circ}}$. Hence,
$\tan \theta = \cot ({90^{\circ}} - \theta )$.