Question

The value of \[\theta \] in between , \[{{0}^{0}}\] and ${{360}^{0}}$ and satisfying the equation $\tan \theta +\dfrac{1}{\sqrt{3}}=0$ is equal to
A. $\theta ={{150}^{0}}$ and ${{300}^{0}}$.
B. \[\theta ={{120}^{0}}\] and ${{300}^{0}}$
C. \[\theta ={{60}^{0}}\] and \[{{240}^{0}}\]
D. $\theta ={{150}^{0}}$ and \[{{330}^{0}}\]

Answer 1

Hint: To find the value of $\theta $, we will first write the trigonometric function on one side of the equation. Then using trigonometric table of values and formula $\tan (-\theta )=-\tan \theta $ we will form an equation in which we will apply the theorem of general equation according to which for all the even multiples of $\dfrac{\pi }{2}$, $\tan x=\tan y$ implies that $x=n\pi +y$ where $n$ is an integer.

Formula Used: $\tan (-\theta )=-\tan \theta $

Complete step by step solution: We are given a trigonometric equation $\tan \theta +\dfrac{1}{\sqrt{3}}=0$ where value of \[\theta \] lies in between the interval \[{{0}^{0}}\] and ${{360}^{0}}$ and we have to derive the value of $\theta $.
We will take the given equation and write the trigonometric function on one side of the equation.
$\begin{align}
  & \tan \theta +\dfrac{1}{\sqrt{3}}=0 \\
 & \tan \theta =-\dfrac{1}{\sqrt{3}}
\end{align}$
We know that $\tan \dfrac{\pi }{6}=\dfrac{1}{\sqrt{3}}$ . So the equation will be,
\[\tan \theta =-\tan \dfrac{\pi }{6}\]
Using formula $\tan (-\theta )=-\tan \theta $in the equation,
\[\tan \theta =\tan \left( -\dfrac{\pi }{6} \right)\]
We will now apply the theorem of general equation of tan,
\[\begin{align}
  & \theta =n\pi +\left( -\dfrac{\pi }{6} \right) \\
 & \theta =n\pi -\dfrac{\pi }{6}
\end{align}\]
As the value of \[\theta \] lies in between the interval \[{{0}^{0}}\] and ${{360}^{0}}$ , we will substitute the value of $n$ as $n=1,2$.
We know that $\pi ={{180}^{0}}$. So,
\[\begin{align}
  & \theta =(1)\pi -\dfrac{\pi }{6} \\
 & \theta =\dfrac{5\pi }{6} \\
 & \theta =\dfrac{5\times {{180}^{0}}}{6} \\
 & \theta ={{150}^{0}}
\end{align}\] , \[\begin{align}
  & \theta =(2)\pi -\dfrac{\pi }{6} \\
 & \theta =\dfrac{11\pi }{6} \\
 & \theta =\dfrac{11\times {{180}^{0}}}{6} \\
 & \theta ={{330}^{0}} \\
\end{align}\]

The value of $\theta $ for the trigonometric equation $\tan \theta +\dfrac{1}{\sqrt{3}}=0$ when \[\theta \] lies in between the interval \[{{0}^{0}}\] and ${{360}^{0}}$ is $\theta ={{150}^{0}}$ and \[{{330}^{0}}\].

Option ‘D’ is correct

Note: The measure of angles can be depicted either in radians and degrees. Radian can be expressed in degrees as $\pi ={{180}^{0}}$.
To change the measure of angle in degrees from radians we have to multiply the angle with $\dfrac{\pi }{180}$ and to convert from radians to degrees we have to multiply the angle with $\dfrac{180}{\pi }$.