Question

If \[\sin \theta =\sqrt{3}\cos \theta \], \[-\pi <\theta <0\] and then $\theta =$
A. $\dfrac{-5\pi }{6}$
B. $\dfrac{-4\pi }{6}$
C. $\dfrac{4\pi }{6}$
D. $\dfrac{5\pi }{6}$

Answer 1

Hint: To find the value of $\theta $ , we will first write the trigonometric function on side of the given equation. Then we will simplify and derive an equation in term of the trigonometric function tan. We will now apply the theorem of the general equation of tan according to which for all the even multiple of $\dfrac{\pi }{2}$ ,$\tan x=\tan y$implies that $x=n\pi +y$ where \[n\in Z\].

Complete step by step solution: We are given a trigonometric equation \[\sin \theta =\sqrt{3}\cos \theta \] when \[-\pi <\theta <0\] and we have to derive the value of $\theta $.
We will take the given equation and put the function on one side of the equation.
\[\begin{align}
  & \sin \theta =\sqrt{3}\cos \theta \\
 & \dfrac{\sin \theta }{\cos \theta }=\sqrt{3} \\
 & \tan \theta =\sqrt{3}
\end{align}\]
We know that $\tan \dfrac{\pi }{3}=\sqrt{3}$. So,
\[\tan \theta =\tan \dfrac{\pi }{3}\]
Applying the theorem here we will get,
$\theta =n\pi +\dfrac{\pi }{3}$.
Because the interval of $\theta $ lies in the interval \[-\pi <\theta <0\], we will put value of $n$ as $n=-1 $.
$\begin{align}
  & \theta =(-1)\pi +\dfrac{\pi }{3} \\
 & \theta =\dfrac{-2\pi }{3}
\end{align}$
We can write the value of $\theta $ derived as $\theta =\dfrac{-4\pi }{6}$ by multiplying and dividing the fraction by $2$.
The value of $\theta $ for the trigonometric equation \[\sin \theta =\sqrt{3}\cos \theta \] when \[-\pi <\theta <0\] is $\theta =\dfrac{-4\pi }{6}$

Option ‘B’ is correct

Note: The trigonometric function tan can be defined as the ratio of the trigonometric function sin to cos. With respect to right angled triangle we can define it as the ratio of perpendicular side of the triangle to the base of the triangle.
The interval given is \[-\pi <\theta <0\] which means that value of $\theta $ will be less than zero and the value of angle will be in negative because $\tan (-\pi )=-\tan \pi $. That is why the value of $n$ we selected to substitute in the general equation is less than zero.