If $\sin \left( {{120}^{0}}-\alpha \right)=\sin \left( {{120}^{0}}-\beta \right)$ , $0 < \alpha ,\beta < \pi $ , then find the relation between $\alpha $ and $\beta $ .

Hint: For solving this question firstly, we will apply the formula for $\sin C-\sin D$ . After that, we will use the results of equations like $\sin \theta =0$ and $\cos \theta =0$ to write the relation between $\alpha $ and $\beta $ in terms of a variable $n$ where $n$ is an integer. Then, we will consider the inequality $0 < \alpha ,\beta < \pi $ to write our final answer.

Complete step-by-step answer:
Given:
It is given that if $\sin \left( {{120}^{0}}-\alpha \right)=\sin \left( {{120}^{0}}-\beta \right)$ , $0 < \alpha ,\beta < \pi $ and we have to find the relation between $\alpha $ and $\beta $ .
Now, before we proceed we should know the following three formulas:
$\begin{align}
  & \sin C-\sin D=2\cos \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{C-D}{2} \right).................\left( 1 \right) \\
 & \sin \theta =0 \\
 & \Rightarrow \theta =n\pi \text{ }\left( n\in I \right).....................................................\left( 2 \right) \\
 & \cos \theta =0 \\
 & \Rightarrow \theta =\left( 2n+1 \right)\dfrac{\pi }{2}\text{ }\left( n\in I \right).........................................\left( 3 \right) \\
\end{align}$
Now, as we have $\sin \left( {{120}^{0}}-\alpha \right)=\sin \left( {{120}^{0}}-\beta \right)$ . Then,
$\begin{align}
  & \sin \left( {{120}^{0}}-\alpha \right)=\sin \left( {{120}^{0}}-\beta \right) \\
 & \Rightarrow \sin \left( \dfrac{2\pi }{3}-\alpha \right)-\sin \left( \dfrac{2\pi }{3}-\beta \right)=0 \\
\end{align}$
Now, we will use the formula from the equation (1). Then,
$\begin{align}
  & \sin \left( \dfrac{2\pi }{3}-\alpha \right)-\sin \left( \dfrac{2\pi }{3}-\beta \right)=0 \\
 & \Rightarrow 2\cos \left( \dfrac{{}^{2\pi }/{}_{3}-\alpha +{}^{2\pi }/{}_{3}-\beta }{2} \right)\sin \left( \dfrac{{}^{2\pi }/{}_{3}-\alpha -{}^{2\pi }/{}_{3}+\beta }{2} \right)=0 \\
 & \Rightarrow 2\cos \left( \dfrac{2\pi }{3}-\dfrac{\left( \alpha +\beta \right)}{2} \right)\sin \left( \dfrac{\beta -\alpha }{2} \right)=0 \\
\end{align}$
Now, from the above result we will have the following two results:
1. $\cos \left( \dfrac{2\pi }{3}-\dfrac{\left( \alpha +\beta \right)}{2} \right)=0$
2. $\sin \left( \dfrac{\beta -\alpha }{2} \right)=0$
Now, at least one of the above results could be true so, we will solve both results separately.
1. $\cos \left( \dfrac{2\pi }{3}-\left( \alpha +\beta \right) \right)=0$ :
Now, using the formula from the equation (3). Then,
$\begin{align}
  & \cos \left( \dfrac{2\pi }{3}-\dfrac{\left( \alpha +\beta \right)}{2} \right)=0=0 \\
 & \Rightarrow \dfrac{2\pi }{3}-\dfrac{\left( \alpha +\beta \right)}{2}=\left( 2n+1 \right)\dfrac{\pi }{2} \\
 & \Rightarrow -\dfrac{\left( \alpha +\beta \right)}{2}=n\pi +\dfrac{\pi }{2}-\dfrac{2\pi }{3} \\
 & \Rightarrow -\dfrac{\left( \alpha +\beta \right)}{2}=n\pi -\dfrac{\pi }{6} \\
 & \Rightarrow \alpha +\beta =\dfrac{\pi }{3}-2n\pi ..............\left( 4 \right) \\
\end{align}$
Now, we got $\alpha +\beta =\dfrac{\pi }{3}-2n\pi $ where $n$ is an integer. But as it is given that $0 < \alpha < \pi $ and $0 < \beta < \pi $ so, we can add inequality $0 < \alpha < \pi $ and $0 < \beta < \pi $ to get the range of values in which $\alpha +\beta $ lies. Then,
$\begin{align}
  & 0 < \alpha < \pi \\
 & 0 < \beta < \pi \\
 & \Rightarrow 0 < \alpha +\beta < 2\pi \\
\end{align}$
Now, we can substitute $\alpha +\beta =\dfrac{\pi }{3}-2n\pi $ from equation (4), in the above inequality. Then,
$\begin{align}
  & 0 < \alpha +\beta < 2\pi \\
 & \Rightarrow 0 < \dfrac{\pi }{3}-2n\pi < 2\pi \\
\end{align}$
Now, we will subtract $\dfrac{\pi }{3}$ from each term in the above inequality. Then,
$\begin{align}
  & 0 < \dfrac{\pi }{3}-2n\pi < 2\pi \\
 & \Rightarrow -\dfrac{\pi }{3} < -2n\pi < 2\pi -\dfrac{\pi }{3} \\
 & \Rightarrow -\dfrac{\pi }{3} < -2n\pi < \dfrac{5\pi }{3} \\
\end{align}$
Now, we will divide each term by $2\pi $ in the above inequality. Then,
$\begin{align}
  & -\dfrac{\pi }{3} < -2n\pi < \dfrac{5\pi }{3} \\
 & \Rightarrow -\dfrac{\pi }{3\times 2\pi } < -n < \dfrac{5\pi }{3\times 2\pi } \\
 & \Rightarrow -\dfrac{1}{6} < -n < \dfrac{5}{6} \\
\end{align}$
Now, we multiply each term by $-1$ so, the inequality sign will be reversed. Then,
$\begin{align}
  & -\dfrac{1}{6} < -n < \dfrac{5}{6} \\
 & \Rightarrow \dfrac{1}{6} > n > -\dfrac{5}{6} \\
 & \Rightarrow -\dfrac{5}{6} < n < \dfrac{1}{6} \\
\end{align}$
Now, from the above result, we can say that $n\in \left( -\dfrac{5}{6},\dfrac{1}{6} \right)$ and as we know that, $n$ is an integer. And between $-\dfrac{5}{6}$ and $\dfrac{1}{6}$ , $0$ is the only integer. So, the value of $n=0$ only for this case and we can put $n=0$ in equation (4). Then,
$\begin{align}
  & \alpha +\beta =\dfrac{\pi }{3}-2n\pi \\
 & \Rightarrow \alpha +\beta =\dfrac{\pi }{3}......................\left( 5 \right) \\
\end{align}$

2. $\sin \left( \dfrac{\beta -\alpha }{2} \right)=0$ :
Now, using the formula from the equation (2). Then,
$\begin{align}
  & \sin \left( \dfrac{\beta -\alpha }{2} \right)=0 \\
 & \Rightarrow \dfrac{\beta -\alpha }{2}=n\pi \\
 & \Rightarrow \beta -\alpha =2n\pi ..............\left( 6 \right) \\
\end{align}$
Now, we got $\beta -\alpha =2n\pi $ where $n$ is an integer. But as it is given that $0 < \alpha < \pi $ so, we multiply each term by $-1$ in $0 < \alpha < \pi $ then, the inequality sign will be reversed. Then,
$\begin{align}
  & 0 < \alpha < \pi \\
 & \Rightarrow -\pi < -\alpha < 0 \\
\end{align}$
Now, as we know that $0 < \beta < \pi $ so, we can add inequality $-\pi < -\alpha < 0$ from the above with inequality $0 < \beta < \pi $ to get the range of values in which $\beta -\alpha $ lies. Then,
$\begin{align}
  & -\pi < -\alpha < 0 \\
 & 0 < \beta < \pi \\
 & \Rightarrow -\pi < \beta -\alpha < \pi \\
\end{align}$
Now, we can substitute $\beta -\alpha =2n\pi $ from equation (6) in the above inequality. Then,
$\begin{align}
  & -\pi < \beta -\alpha < \pi \\
 & \Rightarrow -\pi < 2n\pi < \pi \\
\end{align}$
Now, we divide each term by $2\pi $ in the above inequality. Then,
$\begin{align}
  & -\pi < 2n\pi < \pi \\
 & \Rightarrow \dfrac{-\pi }{2\pi } < n < \dfrac{\pi }{2\pi } \\
 & \Rightarrow -\dfrac{1}{2} < n < \dfrac{1}{2} \\
\end{align}$
Now, from the above result, we can say that $n\in \left( -\dfrac{1}{2},\dfrac{1}{2} \right)$ and as we know that, $n$ is an integer. And between $-\dfrac{1}{2}$ and $\dfrac{1}{2}$ , $0$ is the only integer. So, the value of $n=0$ only for this case and we can put $n=0$ in equation (6). Then,
$\begin{align}
  & \beta -\alpha =2n\pi \\
 & \Rightarrow \beta -\alpha =0 \\
 & \Rightarrow \beta =\alpha ...........................\left( 7 \right) \\
\end{align}$
Now, from the equation (5) and equation (7), we have the following two relations between $\alpha $ and $\beta $ :
1. $\alpha +\beta =\dfrac{\pi }{3}$ .
2. $\beta =\alpha $ .
Thus, at least one of the above relation should be true if $\sin \left( {{120}^{0}}-\alpha \right)=\sin \left( {{120}^{0}}-\beta \right)$ , $0 < \alpha ,\beta < \pi $ .

Note: Here, the student should first try to understand what is asked in the problem. After that, we should apply the formula of $\sin C-\sin D$ correctly and in this question always remember while solving that it is given that $0 < \alpha ,\beta < \pi $ so, we should consider this inequality also to write our final answer. Moreover, always avoid calculation mistakes while solving the problem to get the correct answer.