Question

The most general value of $\theta $ which will satisfy both the equations $\sin \theta =-\dfrac{1}{2}$ and $\tan \theta =\dfrac{1}{\sqrt{3}}$ is
A.$n\pi +{{(-1)}^{n}}\dfrac{\pi }{6}$
B. $n\pi +\dfrac{\pi }{6}$
C. $2n\pi \pm \dfrac{\pi }{6}$
D. None of these.

Answer 1

Hint: We will first find the quadrant in which $\theta $ lies and then find its value accordingly with the help of trigonometric table of values. Using trigonometric table of values at specific angles for both equations we will derive new equations and then apply the theorem of general solution of sin and tan.
The theorem of general solution of sin states that for all the real values of $x$ and $y$, $\sin x=\sin y$ implies that $x=n\pi +{{(-1)}^{n}}y$and theorem of general solution of tan states that for all the even multiples of $\dfrac{\pi }{2}$, $\tan x=\tan y$implies that $x=2n\pi +y$ where $n\in Z$

Complete step by step solution: We are given trigonometric equations $\sin \theta =-\dfrac{1}{2}$ and $\tan \theta =\dfrac{1}{\sqrt{3}}$ and we have to determine the most general value of $\theta $ which will satisfy both the equations.
We can see from the given equations that $\theta $ is negative for sin and positive for tan which means that $\theta $ lies in the third quadrant where angle is represented as $\pi +\theta $.
As we know that $\sin \dfrac{\pi }{6}=\dfrac{1}{2}$ so the equation will be,
$\begin{align}
  & \sin \theta =-\sin \dfrac{\pi }{6} \\
 & \sin \theta =\sin \left( \pi +\dfrac{\pi }{6} \right) \\
 & \sin \theta =\sin \dfrac{7\pi }{6} \\
\end{align}$
Applying theorem of general equation of sin,
$\theta =n\pi +{{(-1)}^{n}}\dfrac{7\pi }{6}$…. (i)
Now we will take equation $\tan \theta =\dfrac{1}{\sqrt{3}}$. We know that $\tan \dfrac{\pi }{6}=\dfrac{1}{\sqrt{3}}$so,
$\begin{align}
  & \tan \theta =\tan \dfrac{\pi }{6} \\
 & \tan \theta =\tan \left( \pi +\dfrac{\pi }{6} \right) \\
 & \tan \theta =\tan \left( \dfrac{7\pi }{6} \right) \\
\end{align}$
Applying the theorem of general solution of tan, we will get,
$\theta =n\pi +\dfrac{7\pi }{6}$ …. (ii)
Now there are two values of $\theta $ satisfying each of the two equations that is $\theta =n\pi +{{(-1)}^{n}}\dfrac{7\pi }{6}$ , $\theta =n\pi +\dfrac{7\pi }{6}$ .
The common value in both the equations (i) and (ii) is $\dfrac{7\pi }{6}$and this value of will repeat after every $2n\pi $ interval so the most general equation will be $\theta =2n\pi +\dfrac{7\pi }{6}$.
The most general value of $\theta $ satisfying the equations $\sin \theta =-\dfrac{1}{2}$ and $\tan \theta =\dfrac{1}{\sqrt{3}}$ is $\theta =2n\pi +\dfrac{7\pi }{6}$.

Option ‘D’ is correct

Note: The trigonometric quadrants have different intervals for each of the four quadrants. The first quadrant has interval of $\left[ 0,\dfrac{\pi }{2} \right]$, second quadrant has $\left[ \dfrac{\pi }{2},\pi \right]$, third quadrant has $\left[ \pi ,\dfrac{3\pi }{2} \right]$ and fourth quadrant has $\left[ \dfrac{3\pi }{2},2\pi \right]$.