Prove that $\sin 10\cdot \sin 30\cdot \sin 50\cdot \sin 70=\dfrac{1}{16}$?

Hint: In the given question, we need to use the trigonometric identities and need to find the value of the given trigonometric equation. Also, we need to know the value of sine function at different angles and need to use only sine trigonometric identities as only that is involved in the question.

Complete step-by-step solution:
In the given question we need to compute the trigonometric equation using various trigonometric identities involving sine function.
Some identities that are to be used in this question are:
$\sin 2A=2\sin A\cos A$
$\sin (90-A)=\cos A$
$\sin 30=\dfrac{1}{2}$
Now, considering the left- hand side of the equation we have,
$\sin 10\cdot \sin 30\cdot \sin 50\cdot \sin 70=\sin 10\cdot \dfrac{1}{2}\cdot \sin 50\cdot \sin 70$
Now, left-hand side implies that
$\begin{align}
  & \sin 10\cdot \dfrac{1}{2}\cdot \sin 50\cdot \sin 70 \\
 & \Rightarrow \dfrac{1}{2}\sin 10\cdot \sin \left( 90-40 \right)\cdot \sin \left( 90-20 \right) \\
 & \Rightarrow \dfrac{1}{2}\sin 10\cdot \cos 40\cdot \cos 20 \\
\end{align}$
Now, we need to multiply with $\dfrac{2\cos 10}{2\cos 10}$ and we get,
$\begin{align}
  & \dfrac{1}{2}\cdot \dfrac{1}{2\left( \cos 10 \right)}\cdot 2\cos 10\cdot \sin 10\cdot \cos 40\cdot \cos 20 \\
 & \Rightarrow \dfrac{1}{4\cos 10}\left( \sin 20\cdot \cos 40\cdot \cos 20 \right) \\
\end{align}$
Now, multiplying and dividing by 2 and using the identity mentioned above we get,
$\begin{align}
  & \dfrac{1}{8\cos 10}\left( 2\sin 20\cdot \cos 20\cdot \cos 40 \right) \\
 & \Rightarrow \dfrac{1}{8\cos 10}\left( \sin 40\cdot \cos 40 \right) \\
 & \Rightarrow \dfrac{1}{16\cos 10}\left( \sin 80 \right) \\
\end{align}$
Now, again using the previous identity and again multiplying and dividing by 2 we get,
 $\begin{align}
  & \dfrac{1}{16\cos 10}\left( \sin \left( 90-10 \right) \right) \\
 & \Rightarrow \dfrac{1}{16\cos 10}\cos 10 \\
 & \Rightarrow \dfrac{1}{16} \\
\end{align}$
And we know that the right-hand side of the equation is also $\dfrac{1}{16}$ .
Therefore, we have proved that the left-hand side and right-hand side of the equation is equal.

Note: For the given question we need to make sure that we apply the correct identity and hence obtain the value. Also, we can convert the sine function into other functions and then attain the value but for this question it will lead to complexity so we need to be careful while operating with the sine angles.