Let ${{\tan }^{-1}}\left( y \right)={{\tan }^{-1}}\left( x \right)+{{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)$, where $\left| x \right|<\dfrac{1}{\sqrt{3}}$, then the value of y is
(a) $\dfrac{3x-{{x}^{3}}}{1-3{{x}^{2}}}$
(b) $\dfrac{3x+{{x}^{3}}}{1-3{{x}^{2}}}$
(c) $\dfrac{3x-{{x}^{3}}}{1+3{{x}^{2}}}$
(d) $\dfrac{3x-{{x}^{3}}}{1+3{{x}^{2}}}$

Hint: To solve this question, we will first list out the properties of tan inverse of composite angles. Then we will choose the formula for ${{\tan }^{-1}}A+{{\tan }^{-1}}B$ and substitute A = x and B = $\dfrac{2x}{1-{{x}^{2}}}$. Thus, we will get tan inverse on both sides of equal to sign. We can then cancel tan inverse on both sides and get the value of y.

Complete step-by-step answer:
To begin with, we will understand the concept of inverse trigonometric rations. If $\tan \theta =\dfrac{a}{b}$, then ${{\tan }^{-1}}\left( \dfrac{a}{b} \right)=\theta $. Thus, if we inverse the trigonometric ratio, its domain changes into range and range changes to domain.
From the chapter of inverse trigonometry, we know the following formulas:
We know from trigonometry that $\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$
Let ${{\tan }^{-1}}x$ = A and ${{\tan }^{-1}}y$ = B
Thus, tan A = x and tan B = y
$\begin{align}
  & \Rightarrow \tan \left( A+B \right)=\dfrac{x+y}{1-xy} \\
 & \Rightarrow A+B={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right) \\
\end{align}$
Therefore, ${{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)......\left( 1 \right)$, given that xy < 1
Now let us consider the right hand side.
The right hand side is ${{\tan }^{-1}}\left( x \right)+{{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)$.
We will apply the formula (1) on the right hand side.
 $\Rightarrow {{\tan }^{-1}}\left( x \right)+{{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)={{\tan }^{-1}}\left( \dfrac{x+\dfrac{2x}{1-{{x}^{2}}}}{1-\dfrac{x\times 2x}{1-{{x}^{2}}}} \right)$
Now, we shall make the denominator of the numerator and denominator same by taking the LCM $1-{{x}^{2}}$. The common denominator from the numerator and denominator will divide out.
$\begin{align}
  & \Rightarrow {{\tan }^{-1}}\left( x \right)+{{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)={{\tan }^{-1}}\left( \dfrac{\dfrac{\left( x \right)\left( 1-{{x}^{2}} \right)+2x}{1-{{x}^{2}}}}{\dfrac{\left( 1-{{x}^{2}} \right)-x\times 2x}{1-{{x}^{2}}}} \right) \\
 & \Rightarrow {{\tan }^{-1}}\left( x \right)+{{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)={{\tan }^{-1}}\left( \dfrac{\left( x \right)\left( 1-{{x}^{2}} \right)+2x}{\left( 1-{{x}^{2}} \right)-2{{x}^{2}}} \right) \\
\end{align}$
Now, we will simplify it further.
$\begin{align}
  & \Rightarrow {{\tan }^{-1}}\left( x \right)+{{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)={{\tan }^{-1}}\left( \dfrac{x-{{x}^{3}}+2x}{1-3{{x}^{2}}} \right) \\
 & \Rightarrow {{\tan }^{-1}}\left( x \right)+{{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)={{\tan }^{-1}}\left( \dfrac{3x-{{x}^{3}}}{1-3{{x}^{2}}} \right)......\left( 3 \right) \\
\end{align}$
We are also given that $\left| x \right|<\dfrac{1}{\sqrt{3}}$
$\begin{align}
  & \Rightarrow {{x}^{2}}<\dfrac{1}{3} \\
 & \Rightarrow 1>3{{x}^{2}} \\
\end{align}$
So, there AB < 1 satisfies.
Therefore, ${{\tan }^{-1}}\left( y \right)={{\tan }^{-1}}\left( \dfrac{3x-{{x}^{3}}}{1-3{{x}^{2}}} \right)$
Thus, $y=\dfrac{3x-{{x}^{2}}}{1-3{{x}^{2}}}$.

So, the correct answer is “Option (a)”.

Note: The steps employed for derivation of the formula ${{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)$ is not required in the solution. That part is included so that students can understand the source of this formula. Students can directly recollect this formula and write it while solving similar problems.