Prove the following statement, $\sin \left[ {{\tan }^{-1}}\left( \dfrac{1-{{x}^{2}}}{2x} \right)+{{\cos }^{-1}}\left( \dfrac{1-{{x}^{2}}}{1+{{x}^{2}}} \right) \right]=1$.

Hint:In order to solve this question, we should have some knowledge of the inverse trigonometric formulas like, ${{\cos }^{-1}}\left( \dfrac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)=2{{\tan }^{-1}}x$ and $2{{\tan }^{-1}}x={{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)$ and ${{\tan }^{-1}}x={{\cot }^{-1}}\dfrac{1}{x}$ and ${{\cot }^{-1}}x+{{\tan }^{-1}}x=\dfrac{\pi }{2}$. By using these formulas, we can prove the desired relation.

Complete step-by-step answer:
In the given question, we have been asked to prove that, $\sin \left[ {{\tan }^{-1}}\left( \dfrac{1-{{x}^{2}}}{2x} \right)+{{\cos }^{-1}}\left( \dfrac{1-{{x}^{2}}}{1+{{x}^{2}}} \right) \right]=1$. To prove the same, we will first consider the left hand side or the LHS of the equality. So, we can write the LHS as,
$LHS=\sin \left[ {{\tan }^{-1}}\left( \dfrac{1-{{x}^{2}}}{2x} \right)+{{\cos }^{-1}}\left( \dfrac{1-{{x}^{2}}}{1+{{x}^{2}}} \right) \right]$
Now, we know that, ${{\cos }^{-1}}\left( \dfrac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)=2{{\tan }^{-1}}x$. So, we can apply it here in the above equality and get the LHS as follows,
$LHS=\sin \left[ {{\tan }^{-1}}\left( \dfrac{1-{{x}^{2}}}{2x} \right)+2{{\tan }^{-1}}x \right]$
Now, we know that $2{{\tan }^{-1}}x$ can be expressed as ${{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)$. So, we will substitute it in the above equality. So, we get the LHS as,
$LHS=\sin \left[ {{\tan }^{-1}}\left( \dfrac{1-{{x}^{2}}}{2x} \right)+{{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right) \right]$
Now, we also know that ${{\tan }^{-1}}x={{\cot }^{-1}}\dfrac{1}{x}$. So, we can write ${{\tan }^{-1}}\left( \dfrac{1-{{x}^{2}}}{2x} \right)$ as ${{\cot }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)$. Therefore, we can substitute it in the LHS. So, we get,
$LHS=\sin \left[ {{\cot }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)+{{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right) \right]$
Now, we know that ${{\cot }^{-1}}x+{{\tan }^{-1}}x=\dfrac{\pi }{2}$. So, by using this identity, we can write the LHS as,
$LHS=\sin \left( \dfrac{\pi }{2} \right)$
We know that the sine ratio of $\dfrac{\pi }{2}$ is 1, that is $\sin \left( \dfrac{\pi }{2} \right)=1$. So, we can write the LHS as,
$LHS=1$.
We can see that it is the same as the right hand side or the RHS of the given equality. Therefore LHS = RHS.
Hence, we have proved the given equality.

Note: While solving the given question, one can think of applying the identity of ${{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)$ in the step, where we applied ${{\tan }^{-1}}x={{\cot }^{-1}}\dfrac{1}{x}$. This would not be incorrect, but it is a lengthy method, so it is better to avoid this and leave the solution simpler.