Question

How do you evaluate $\cot \left( {\dfrac{{5\pi }}{6}} \right)$?

Answer 1

Hint: In order to determine the value of the above question, first rewriting the angle in the form of $\pi \pm \theta $, we get $\pi - \dfrac{\pi }{6}$ where $\theta = \dfrac{\pi }{6}$. As we know that $\cot (\pi - \theta ) = - \cot (\theta )$ because $\pi - \dfrac{\pi }{6}$ is the angle in the second quadrant and cotangent is always negative in $2^{nd}$ quadrant. Rewrite the cotangent and put the exact value of $\cot \left( {\dfrac{\pi }{6}} \right) = \sqrt 3 $ to get your required result.

Complete step by step answer:
We are given a $\cot \left( {\dfrac{{5\pi }}{6}} \right)$, and we have to evaluate its value.
Let’s write the angle $\dfrac{{5\pi }}{6}$in the form of $\pi \pm n$. We get that $\dfrac{{5\pi }}{6}$ can be written as $\pi - \dfrac{\pi }{6}$
$ = \cot \left( {\pi - \dfrac{\pi }{6}} \right)$-----(1)
Note that $\cot (\pi - \theta ) = - \cot (\theta )$
As we can see that $\pi - \dfrac{\pi }{6}$ is the angle in the second quadrant and cotangent is always negative in $2^{nd}$ quadrant, that’s by $\cot (\pi - \theta ) = - \cot (\theta )$.
We can write
$
   = \cot \left( {\pi - \dfrac{\pi }{6}} \right) \\
   = - \cot \left( {\dfrac{\pi }{6}} \right) \\
 $
The exact value of $\cot \left( {\dfrac{\pi }{6}} \right) = \sqrt 3 $, substituting this value, we get
$ = - \sqrt 3 $

Therefore, the value of $\cot \left( {\dfrac{{5\pi }}{6}} \right)$ is equal to $ - \sqrt 3 $.

Additional information:
1. Periodic Function= A function $f(x)$ is said to be a periodic function if there exists a real number T > 0 such that $f(x + T) = f(x)$ for all x.
If T is the smallest positive real number such that $f(x + T) = f(x)$ for all x, then T is called the fundamental period of $f(x)$.
Since $\sin \,(2n\pi + \theta ) = \sin \theta $ for all values of $\theta $ and n$ \in $N.
2. Even Function – A function $f(x)$ is said to be an even function,if $f( - x) = f(x)$ for all x in its domain.
Odd Function – A function $f(x)$ is said to be an even function,if $f( - x) = - f(x)$ for all x in its domain.
We know that $\sin ( - \theta ) = - \sin \theta.\cos ( - \theta ) = \cos \theta \,and\,\tan ( - \theta ) = - \tan \theta $
Therefore, $\sin \theta $ and $\tan \theta $ and their reciprocals,$\cos ec\theta $ and $\cot \theta $ are odd functions whereas \[\cos \theta \] and its reciprocal \[\sec \theta \] are even functions.
3. Trigonometry is one of the significant branches throughout the entire existence of mathematics and this idea is given by a Greek mathematician Hipparchus.

Note: 1. One must be careful while taking values from the trigonometric table and cross-check at least once to avoid any error in the answer.
2. Cotangent and tangent are always negative in $2^{nd}$ quadrant and positive in $1^{st}$ and $3^{rd}$ quadrant.