Write the value of ${{\sin }^{-1}}\left( \dfrac{1}{3} \right)-{{\cos }^{-1}}\left( -\dfrac{1}{3} \right)$.

Hint:We will apply the basic identity which is used in inverse trigonometric functions. This formula is given by ${{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2}$. Because of this we will find the value of ${{\cos }^{-1}}\left( -\dfrac{1}{3} \right)$ in the expression ${{\sin }^{-1}}\left( \dfrac{1}{3} \right)-{{\cos }^{-1}}\left( -\dfrac{1}{3} \right)$.

Complete step-by-step answer:
We will consider the expression ${{\sin }^{-1}}\left( \dfrac{1}{3} \right)-{{\cos }^{-1}}\left( -\dfrac{1}{3} \right)...(i)$.
Now, to solve this expression we need to use the basic identity of inverse trigonometric functions. This is given by ${{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2}$. This can also be written as ${{\cos }^{-1}}x=\dfrac{\pi }{2}-{{\sin }^{-1}}x$ by placing the inverse sign term to the right side of the equation. In this formula we will substitute the value of x as $-\dfrac{1}{3}$. Therefore, we have ${{\cos }^{-1}}\left( -\dfrac{1}{3} \right)=\dfrac{\pi }{2}-{{\sin }^{-1}}\left( -\dfrac{1}{3} \right)$.
Now, we will substitute the value of ${{\cos }^{-1}}\left( -\dfrac{1}{3} \right)$ in the expression (i). Thus, our expression is converted into a new expression which is given by ${{\sin }^{-1}}\left( \dfrac{1}{3} \right)-{{\cos }^{-1}}\left( -\dfrac{1}{3} \right)={{\sin }^{-1}}\left( \dfrac{1}{3} \right)-\left( \dfrac{\pi }{2}-{{\sin }^{-1}}\left( -\dfrac{1}{3} \right) \right)$.
After multiplying the minus sign in the equation we will get ${{\sin }^{-1}}\left( \dfrac{1}{3} \right)-{{\cos }^{-1}}\left( -\dfrac{1}{3} \right)={{\sin }^{-1}}\left( \dfrac{1}{3} \right)-\dfrac{\pi }{2}+{{\sin }^{-1}}\left( -\dfrac{1}{3} \right)$.
At this step we will use the formula which is given by $\sin \left( -x \right)=-\sin \left( x \right)$ where x can be any integer.
Therefore, now we after substituting x as $-\dfrac{1}{3}$ again we will have $\sin \left( -\dfrac{1}{3} \right)=-\sin \left( \dfrac{1}{3} \right)$.
Now, we will substitute the value $\sin \left( -\dfrac{1}{3} \right)=-\sin \left( \dfrac{1}{3} \right)$ in ${{\sin }^{-1}}\left( \dfrac{1}{3} \right)-{{\cos }^{-1}}\left( -\dfrac{1}{3} \right)={{\sin }^{-1}}\left( \dfrac{1}{3} \right)-\dfrac{\pi }{2}+{{\sin }^{-1}}\left( -\dfrac{1}{3} \right)$. Therefore, we get
$\begin{align}
  & {{\sin }^{-1}}\left( \dfrac{1}{3} \right)-{{\cos }^{-1}}\left( -\dfrac{1}{3} \right)={{\sin }^{-1}}\left( \dfrac{1}{3} \right)-\dfrac{\pi }{2}+{{\sin }^{-1}}\left( -\dfrac{1}{3} \right) \\
 & \Rightarrow {{\sin }^{-1}}\left( \dfrac{1}{3} \right)-{{\cos }^{-1}}\left( -\dfrac{1}{3} \right)={{\sin }^{-1}}\left( \dfrac{1}{3} \right)-\dfrac{\pi }{2}+\left( -{{\sin }^{-1}}\left( \dfrac{1}{3} \right) \right) \\
 & \Rightarrow {{\sin }^{-1}}\left( \dfrac{1}{3} \right)-{{\cos }^{-1}}\left( -\dfrac{1}{3} \right)={{\sin }^{-1}}\left( \dfrac{1}{3} \right)-\dfrac{\pi }{2}-{{\sin }^{-1}}\left( \dfrac{1}{3} \right) \\
\end{align}$
Now we will cancel both inverse sine terms which are in the right side of the equation with negative and positive sign. This results into
$\begin{align}
  & {{\sin }^{-1}}\left( \dfrac{1}{3} \right)-{{\cos }^{-1}}\left( -\dfrac{1}{3} \right)={{\sin }^{-1}}\left( \dfrac{1}{3} \right)-\dfrac{\pi }{2}-{{\sin }^{-1}}\left( \dfrac{1}{3} \right) \\
 & \Rightarrow {{\sin }^{-1}}\left( \dfrac{1}{3} \right)-{{\cos }^{-1}}\left( -\dfrac{1}{3} \right)=-\dfrac{\pi }{2} \\
\end{align}$
Thus, the required value of the expression ${{\sin }^{-1}}\left( \dfrac{1}{3} \right)-{{\cos }^{-1}}\left( -\dfrac{1}{3} \right)=-\dfrac{\pi }{2}$.

Note: We should always keep this in mind that whenever we take any term to the either side of the equation we have to take care that it should always be converted into its opposite sign. For example in this question we have taken the inverse sine term to the right side of the expression and we have changed the positive sign to the negative sign and then pursued it with the formula.