What is $\sin 0$ divided by $\sin 0$?

Hint: To solve this type of question we need to know the concept of trigonometry and the value of the trigonometry that would tend for a certain angle. We should also have a knowledge of indeterminate form which means the unknown value. We will see some chances where there are some limits which do exist for the function $\dfrac{\sin 0}{\sin 0}$ .

Complete step by step answer:
The question asks us to find the value when $\sin 0$ (sin of zero Radian) which is divided by (sin of zero radian). Question in mathematical form is written as
$\Rightarrow \dfrac{\sin 0}{\sin 0}$
To solve further we need to know the value of $\sin 0$, so the value of $\sin 0$ is $0$. Substituting the same value in the sin function we get:
$\Rightarrow \dfrac{0}{0}$
$\therefore $ The form $\dfrac{0}{0}$ is one of the indeterminate forms, which means there is no certain value for the given question.

Note: A fact that needs to be known is that there are in total seven indeterminate forms and $\dfrac{0}{0}$ is one of them. Division of any number with zero is undefined or could be said as it results in infinity. On considering the above question there are some limits where the function $\dfrac{\sin 0}{\sin 0}$ exists. Let us check the place where the function exist,
$\displaystyle \lim_{x \to 0}\dfrac{\sin x}{\sin x}$
On dividing both the sin function with $x$we get:
$\Rightarrow \displaystyle \lim_{x \to 0}\dfrac{\left( \dfrac{\sin x}{x} \right)}{\left( \dfrac{\sin x}{x} \right)}$
We know that the $\displaystyle \lim_{x \to 0}\dfrac{\sin x}{x}$ is equal to $1$, so substituting this in the above equation we get:
$\Rightarrow \displaystyle \lim_{x \to 0}\dfrac{1}{1}$
As we can see there is no term of $x$in the above equation. In the above equation $1$ is divided by $1$ and we know that any number divided by $1$ gives the number itself, so the value we got is $1$.
So it could be said that at some limits of $x$ the function exists.