Question

Find the value of trigonometric expression $\tan \dfrac{11\pi }{6}$ .

Answer 1

Hint:Convert the given angle in the expression of the problem to acute angle form i.e. between ${{0}^{\circ }}$ to ${{90}^{\circ }}$ (or $\dfrac{\pi }{2}$ ). And use the quadrant rules that tan function is positive in the first and third quadrant and negative in the second and fourth quadrant. If the angle in tan function has involvement of multiple of $\dfrac{\pi }{2}$ (not $\pi $) i.e. of type $\dfrac{n\pi }{2}\pm \theta $ , then change tan function to cot or if summation of angle is of type $n\pi \pm \theta $ , then do not change the trigonometric function. Use the above rules to get the answer. Use $:\tan \left( \dfrac{\pi }{6} \right)=\dfrac{1}{\sqrt{3}}$ .

Complete step-by-step answer:
The expression in the problem is $=\tan \left( \dfrac{11\pi }{6} \right)$ .
Let us suppose the value of the given expression is ‘M’, So, we can write equation as –
$M=\tan \left( \dfrac{11\pi }{6} \right)$ ………………………….. (i)
Now, as we can observe that the angle involved in the expression is $\dfrac{11\pi }{6}$ ,
So, we need to convert the angle to acute angle form i.e. between 0 to $\dfrac{\pi }{2}$ , so that we will be able to get the exact value of that expression.
Now, we can write the given angle in the expression i.e. $\dfrac{11\pi }{6}$ as –
$\dfrac{11\pi }{6}=\left( 2\pi -\dfrac{\pi }{6} \right)$ ………………… (ii)
Hence, the equation (i) can be written with the help of above equation as-
$M=\left( 2\pi -\dfrac{\pi }{6} \right)$ ……………………… (iii)
Now, we know the angle $\left( 2\pi -\dfrac{\pi }{6} \right)$ will lie in the fourth quadrant and as per the rules of trigonometric function in four quadrants, tan function is negative in fourth quadrant. And as the difference of angle written in terms of multiple of $\pi \left( 2\pi \right)$ , it means the trigonometric function will remain the same. And hence, the trigonometric relation for $\tan \left( 2\pi -\theta \right)$ can be given as –
$\tan \left( 2\pi -\theta \right)=-\tan \theta $ ………………… (iv)
Hence, using equation (iv), we get the value of the equation (iii) as –
$M=\tan \left( 2\pi -\dfrac{\pi }{6} \right)=-\tan \dfrac{\pi }{6}$ …………………………………… (v)
Now, as we know the value of $\tan \dfrac{\pi }{6}$ or $\tan {{30}^{\circ }}$ can be given as –
$\tan \dfrac{\pi }{6}=\dfrac{1}{\sqrt{3}}$ …………………….. (vi)
Now, we can put value of $\tan \dfrac{\pi }{6}$ as $\dfrac{1}{\sqrt{3}}$ from the above equation (v) to get the value of given expression in the problem i.e. M. Hence, we get –
$M=-\dfrac{1}{\sqrt{3}}$
So, we get the value of $\tan \left( \dfrac{11\pi }{6} \right)$ as $-\dfrac{1}{\sqrt{3}}$ .
Hence, $-\dfrac{1}{\sqrt{3}}$ is the answer of the given problem.

Note: We need to know two important rules involved for conversion of trigonometric function with respect to the angles.
(i) Take care of the sign with the help of the given trigonometric function and the quadrant in which the angle is lying. This rule can be given as.
(ii) If the angle involved inside trigonometric function is multiply of $\dfrac{\pi }{2}$ (not multiple of $\pi $) i.e. $\dfrac{n\pi }{2}\pm \theta $ type, where n is an odd integer, then we need to convert the
$\begin{align}
  & \sin \rightleftarrows \cos \\
 & \tan \rightleftarrows \cot \\
 & \sec \rightleftarrows \cos ec \\
\end{align}$
And if the angle involved in sum is multiple of $\pi $ i.e. $n\pi \pm \theta $ type, then the trigonometric function will remain the same.
Use the above two rules for conversion of any trigonometric function by changing their angles.
So, one may split the angle $\dfrac{11\pi }{6}$ in terms of $\dfrac{\pi }{2}$ as well, as \[\dfrac{3\pi }{2}+\dfrac{\pi }{3}\].So, it can be another approach but need to apply trigonometric rules for conversion according to it as well.As it is multiple of $\dfrac{\pi }{2}$ tan value changes to cot and in 4th quadrant tan will be negative so $\cot\left(\dfrac{3\pi }{2}+\dfrac{\pi }{3}\right)$=${\cot\left(-\dfrac{\pi}{3}\right)}$=$-\dfrac{1}{\sqrt{3}}$.