If we have a trigonometric expression as $\sin \theta =\operatorname{Sin}\alpha $ , then:
A. $\dfrac{\theta +\alpha }{2}$ is any odd multiple of $\dfrac{\pi }{2}$ and $\dfrac{\theta -\alpha }{2}$ is any multiple of $\pi $ .
B. $\dfrac{\theta +\alpha }{2}$is any even multiple of $\dfrac{\pi }{2}$ and $\dfrac{\theta -\alpha }{2}$ is any odd multiple of $\pi $ .
C. $\dfrac{\theta +\alpha }{2}$is any multiple of $\dfrac{\pi }{2}$ and $\dfrac{\theta -\alpha }{2}$ is any odd multiple of $\pi $ .
D. $\dfrac{\theta +\alpha }{2}$is any multiple of $\dfrac{\pi }{2}$ and $\dfrac{\theta -\alpha }{2}$ is any even multiple of $\pi $ .

Hint: Take $\sin \alpha $ to LHS and then apply the formula “$\sin C-\sin D=2\sin \left( \dfrac{C-D}{2} \right)\cos \left( \dfrac{C+D}{2} \right)$“. In the obtained equation to get $2\sin \left( \dfrac{\theta -\alpha }{2} \right)\cos \left( \dfrac{\theta +\alpha }{2} \right)=0$. Now solve this equation to get values of $\dfrac{\theta -\alpha }{2}$ and $\dfrac{\theta +\alpha }{2}$ .

Complete step-by-step solution:
Given $\sin \theta =\operatorname{Sin}\alpha $
Taking all the terms to LHS, we will get,
$\sin \theta -\sin \alpha =0$ ………………………. (1)
We know that: $\sin C-\sin D=2\sin \left( \dfrac{C-D}{2} \right)\cos \left( \dfrac{C+D}{2} \right)$.
Using this formula in equation (1), we will get,
 $2\sin \left( \dfrac{\theta -\alpha }{2} \right)\cos \left( \dfrac{\theta +\alpha }{2} \right)=0$ .
Dividing both sides of the equation by 2, we will get,
$\Rightarrow \sin \left( \dfrac{\theta -\alpha }{2} \right)\cos \left( \dfrac{\theta +\alpha }{2} \right)=0$
Either $\sin \left( \dfrac{\theta -\alpha }{2} \right)=0$ or $\cos \left( \dfrac{\theta +\alpha }{2} \right)=0$ .
Let us first solve $\sin \left( \dfrac{\theta -\alpha }{2} \right)=0$ .
We know that $\sin \left( 0 \right)=0$ .
So, $\sin \left( \dfrac{\theta -\alpha }{2} \right)=\sin \left( 0 \right)$ .
We know that the general solution of $\sin y=\sin x$ is $x=n\pi +{{\left( -1 \right)}^{n}}y$ .
On putting $x=\dfrac{\left( \theta -\alpha \right)}{2}$ and $y=0$ we will get –
$\Rightarrow \dfrac{\theta -\alpha }{2}=n\pi -0$
So, $ \dfrac{\theta -\alpha }{2}= n\pi $
Now let us solve $\cos \left( \dfrac{\theta +\alpha }{2} \right)=0$ .
We know that \[\cos \left( \dfrac{\pi }{2} \right)=0\] .
So, $\cos \left( \dfrac{\theta +\alpha }{2} \right)=\cos \left( \dfrac{\pi }{2} \right)$ .
We know that the general solution of $\cos y=\cos x$ is $y=x+2n\pi $ .
Where ‘n’ is an integer.
So, general solution of $\cos \left( \dfrac{\theta +\alpha }{2} \right)=\cos \left( \dfrac{\pi }{2} \right)$ is
\[\begin{align}
  & \dfrac{\theta +\alpha }{2}=\dfrac{\pi }{2}+2n\pi \\
 & \Rightarrow \dfrac{\theta -\alpha }{2}=\dfrac{\pi }{2}+2n\pi \\
\end{align}\]
Taking $\left( \dfrac{\pi }{2} \right)$ in RHS, we will get,
$\Rightarrow \dfrac{\theta +\alpha }{2}=\dfrac{\pi }{2}\left( 1+4n \right)$
So, for equation (1) to hold,
Two possible cases,
Case: 1 $\sin \left( \dfrac{\theta -\alpha }{2} \right)=0$ .
From above solution, we have got,
 $\sin \left( \dfrac{\theta -\alpha }{2} \right)=0\Rightarrow \dfrac{\theta -\alpha }{2}=n\pi $ .
i.e. $\dfrac{\theta -\alpha }{2}$ is a multiple of $\pi $ .
Case: 2 $\cos \left( \dfrac{\theta +\alpha }{2} \right)=0$ .
Form above solution, we have got that,
$\cos \left( \dfrac{\theta +\alpha }{2} \right)=0\Rightarrow \dfrac{\theta +\alpha }{2}=\dfrac{\pi }{2}\left( 4n+1 \right)$ .
i.e. $\dfrac{\theta +\alpha }{2}$ is an odd multiple of $\dfrac{\pi }{2}$ .
Hence option (A) is the correct answer.

Note: In the solution, we have used the general solution of $\sin y=\sin x$ is $x=n\pi +{{\left( -1 \right)}^{n}}y$
We can prove this as follows:
$\sin x=\sin y$
So, $x=2k\pi +y$ or $x=2k\pi +\pi -y$
$\Rightarrow x=2k\pi +y$ or $x=\left( 2k+1 \right)\pi -y$
$\Rightarrow x=2k\pi +\left( -1 \right)2ky$ or $x=\left( 2k+1 \right)\pi +\left( -1 \right)2k+1y$
$\Rightarrow x=n\pi +\left( -1 \right)ny$
Where, n is any integer i.e. $n=0,\pm 1,\pm 2,\pm 3............$