The value of the trigonometric expression $\sin \left( 2{{\cos }^{-1}}\left( -\dfrac{3}{5} \right) \right)$ is,
(a) $\dfrac{6}{25}$
(b) $\dfrac{24}{25}$
(c) $\dfrac{4}{5}$
(d) $-\dfrac{24}{25}$

Hint:We will use the basic trigonometric identity which is given by ${{\sin }^{2}}r+{{\cos }^{2}}r=1$ in order to get the value of the expression $\sin \left( 2{{\cos }^{-1}}\left( -\dfrac{3}{5} \right) \right)$. After that we will use substitution in the trigonometric terms.

Complete step-by-step answer:
Now, we will first consider the expression $\sin \left( 2{{\cos }^{-1}}\left( -\dfrac{3}{5} \right) \right)...(i)$.
To solve it we will start by substituting ${{\cos }^{-1}}\left( -\dfrac{3}{5} \right)$ as equal to r. Therefore, we have ${{\cos }^{-1}}\left( -\dfrac{3}{5} \right)=r$.
By taking the inverse cosine term to the right side of the equal sign we will get $\cos r=-\dfrac{3}{5}$.
At this step we will apply the formula of a basic trigonometric identity which is given by ${{\sin }^{2}}r+{{\cos }^{2}}r=1$. Thus, we will have
$\sin r=\sqrt{1-{{\left( \cos r \right)}^{2}}}$
After substituting the value of $\cos r=-\dfrac{3}{5}$ in $\sin r=\sqrt{1-{{\left( \cos r \right)}^{2}}}$. Therefore, we have $\sin r=\sqrt{1-{{\left( -\dfrac{3}{5} \right)}^{2}}}$ which further results into $\sin r=\dfrac{4}{5}$.
Now we will again consider the expression $\sin \left( 2{{\cos }^{-1}}\left( -\dfrac{3}{5} \right) \right)$ and substitute the value ${{\cos }^{-1}}\left( -\dfrac{3}{5} \right)=r$ in it. Thus, we have a new expression which is written as
$\sin \left( 2{{\cos }^{-1}}\left( -\dfrac{3}{5} \right) \right)=\sin \left( 2r \right)$.
Now, we will use the formula of double angle which is given as $\sin \left( 2r \right)=2\sin r\cos r$. Therefore, we get
$\begin{align}
  & \sin \left( 2{{\cos }^{-1}}\left( -\dfrac{3}{5} \right) \right)=\sin \left( 2r \right) \\
 & \Rightarrow \sin \left( 2r \right)=2\sin r\cos r \\
\end{align}$
Now we will do the substitution here. We will put $\sin r=\dfrac{4}{5}$ and $\cos r=-\dfrac{3}{5}$. Therefore, we will have
$\begin{align}
  & \sin \left( 2r \right)=2\sin r\cos r \\
 & \Rightarrow \sin \left( 2r \right)=2\times \left( \dfrac{4}{5} \right)\times \left( -\dfrac{3}{5} \right) \\
\end{align}$
After multiplying the fractions we get $\sin \left( 2r \right)=-\dfrac{24}{25}$.
Therefore, the value of the expression is given by $\sin \left( 2r \right)=-\dfrac{24}{25}$.
Hence, the correct option is (d).

Note: Alternatively we could have solved it by using the formula of $\sin r=\dfrac{\text{perpendicular}}{\text{Hypotenuse}}$ to find the value of $\cos r$ by finding the value of base using the Pythagoras theorem. While taking the inverse trigonometric term to the right side of the equal sign we need to take into consideration the fact that it should be converted into simple trigonometric terms. For example in this question we substituted ${{\cos }^{-1}}\left( -\dfrac{3}{5} \right)=r$ and after by taking the inverse cosine term to the right side of equal sign we converted it into simple cosine function which is written as $\cos r=-\dfrac{3}{5}$.