If $ \sin \left[ {{\cot }^{-1}}\left( x+1 \right) \right]=\cos \left( {{\tan }^{-1}}x \right), $ then \[x=\]
A. $ \dfrac{-1}{2} $
B. $ \dfrac{1}{2} $
C. $ 0 $
D. $ \dfrac{9}{4} $

Hint: We will assume the interior value of $ \sin $ and $ \cos $ present in the LHS and RHS side of equation some variable which will help us to have somewhat simpler equation compared to know, and then we will use different trigonometric identities to find the value of ‘x’.

Complete step by step answer:
Moving ahead with the question in step wise manner;
Let us assume the value of $ {{\cot }^{-1}}\left( x+1 \right)=p $ and $ {{\tan }^{-1}}x=q $ , so we will get new question that is; \[\sin p=\cos q,\]
So as we had assumed $ {{\cot }^{-1}}\left( x+1 \right)=p $ so by inverse trigonometric function we can write it as
 $ \begin{align}
  & {{\cot }^{-1}}\left( x+1 \right)=p \\
\end{align} $
 $ x+1=\cot p $---- equation (i)
Similarly we can write $ {{\tan }^{-1}}x=q $ as;
 $ \begin{align}
  & {{\tan }^{-1}}x=q \\
\end{align} $
 $ x=\tan q $--- equation (ii)
From equation (i) we have $ x+1=\cot p $ , as $ \cot p=\dfrac{B}{P} $ so by comparing we can say that ‘Base’ is $ x+1 $ and perpendicular is 1. So in the modified question we have \[\sin p=\cos q,\] in the LHS side of equation we have \[\sin p\] and by trigonometric identity we know that \[\sin p\] is equal to perpendicular upon hypotenuse, so we can say that \[\sin p=\dfrac{P}{H}\] which will be equal to;
 \[\begin{align}
  & \sin p=\dfrac{P}{H} \\
 & \sin p=\dfrac{1}{\sqrt{{{\left( x+1 \right)}^{2}}+{{1}^{2}}}} \\
\end{align}\]
Similarly solving in the RHS side of equation we have \[\cos q\] , so from equation (ii) we have $ x=\tan q $ , as $ \tan q=\dfrac{P}{B} $ so by comparing we can say that ‘Base’ is 1 and perpendicular is $ x $ . So in the modified question we have \[\sin p=\cos q,\] in the RHS side of equation we have \[\cos q\] and by trigonometric identity we know that \[\cos q\] is equal to Base upon hypotenuse, so we can say that $ \cos q=\dfrac{B}{H} $ which will be equal to;
 $ \begin{align}
  & \cos q=\dfrac{B}{H} \\
 & \cos q=\dfrac{1}{\sqrt{1+{{x}^{2}}}} \\
\end{align} $
So put the value of \[\sin p\] and \[\cos q\] in the modified question we have;
 \[\begin{align}
  & \sin p=\cos q \\
 & \dfrac{1}{\sqrt{{{\left( x+1 \right)}^{2}}+{{1}^{2}}}}=\dfrac{1}{\sqrt{1+{{x}^{2}}}} \\
\end{align}\]
Now simplify the equation to get the value of ‘x’, so first square both sides and cross multiply them, which will give us;
 \[\begin{align}
  & \left( 1+{{x}^{2}} \right)=\left( {{\left( x+1 \right)}^{2}}+{{1}^{2}} \right) \\
 & 1+{{x}^{2}}=1+{{x}^{2}}+2x+1 \\
 & 2x=-1 \\
 & x=-\dfrac{1}{2} \\
\end{align}\]
So we got \[x=-\dfrac{1}{2}\] .

So, the correct answer is “Option A”.

Note: We had assumed the inverse trigonometric function present in the question to some variable with only motive to ease our calculation. Otherwise we can solve them directly without assuming them, then the answer will be the same.