The value of ${{\cos }^{-1}}\left\{ \cos 2{{\cot }^{-1}}\left( \sqrt{2}-1 \right) \right\}$ is equal to
(a) $\sqrt{2}-1$
(b) $\dfrac{\pi }{4}$
(c) $\dfrac{3\pi }{4}$
(d) $0$

Hint: In inverse trigonometric functions, we have a formula ${{\cos }^{-1}}\left( \cos x \right)=x$ if $x$ is a principle angle i.e. $x\in \left[ 0,\pi \right]$. In this question, we will start from the innermost term and convert them to $\cos $ or ${{\cos }^{-1}}$ functions and then use the above formula.

Before proceeding with the question, we must know all the formulas that will be required to solve this question.
In the inverse trigonometric functions, we have the following formulas,
$\left( 1 \right){{\cos }^{-1}}\left( \cos x \right)=x$
$\left( 2 \right){{\cot }^{-1}}x={{\tan }^{-1}}\dfrac{1}{x}$
$\left( 3 \right)2{{\tan }^{-1}}x={{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)$
In the question, we are required to solve ${{\cos }^{-1}}\left\{ \cos 2{{\cot }^{-1}}\left( \sqrt{2}-1 \right) \right\}$. To solve this, we will start from the innermost function and apply the above listed formulas till we reach the outermost function. We will convert all the functions in the form of $\cos $ or ${{\cos }^{-1}}$ with the use of the above listed formulas since the outermost function is a ${{\cos }^{-1}}$ function.
The innermost function is $2{{\cot }^{-1}}\left( \sqrt{2}-1 \right)$. Using formula $\left( 2 \right)$, we get $2{{\cot }^{-1}}\left( \sqrt{2}-1 \right)$ equal to,
 $2{{\cot }^{-1}}\left( \sqrt{2}-1 \right)=2{{\tan }^{-1}}\dfrac{1}{\sqrt{2}-1}$
Using formula $\left( 3 \right)$, we can write \[2{{\tan }^{-1}}\dfrac{1}{\sqrt{2}-1}\] as,
\[\begin{align}
  & 2{{\tan }^{-1}}\dfrac{1}{\sqrt{2}-1}={{\tan }^{-1}}\left( \dfrac{\dfrac{2}{\sqrt{2}-1}}{1-{{\left( \dfrac{1}{\sqrt{2}-1} \right)}^{2}}} \right) \\
 & \Rightarrow 2{{\tan }^{-1}}\dfrac{1}{\sqrt{2}-1}={{\tan }^{-1}}\left( \dfrac{\dfrac{2}{\sqrt{2}-1}}{\dfrac{{{\left( \sqrt{2}-1 \right)}^{2}}-1}{{{\left( \sqrt{2}-1 \right)}^{2}}}} \right) \\
 & \Rightarrow 2{{\tan }^{-1}}\dfrac{1}{\sqrt{2}-1}={{\tan }^{-1}}\left( \dfrac{\dfrac{2}{\sqrt{2}-1}}{\left( \dfrac{2+1-2\sqrt{2}-1}{{{\left( \sqrt{2}-1 \right)}^{2}}} \right)} \right) \\
 & \Rightarrow 2{{\tan }^{-1}}\dfrac{1}{\sqrt{2}-1}={{\tan }^{-1}}\left( \dfrac{\dfrac{2}{\sqrt{2}-1}}{\left( \dfrac{3-2\sqrt{2}-1}{{{\left( \sqrt{2}-1 \right)}^{2}}} \right)} \right) \\
 & \Rightarrow 2{{\tan }^{-1}}\dfrac{1}{\sqrt{2}-1}={{\tan }^{-1}}\left( \dfrac{\dfrac{2}{\sqrt{2}-1}}{\dfrac{2-2\sqrt{2}}{{{\left( \sqrt{2}-1 \right)}^{2}}}} \right) \\
 & \Rightarrow 2{{\tan }^{-1}}\dfrac{1}{\sqrt{2}-1}={{\tan }^{-1}}\left( \dfrac{\dfrac{2}{\sqrt{2}-1}}{\dfrac{-2\left( \sqrt{2}-1 \right)}{{{\left( \sqrt{2}-1 \right)}^{2}}}} \right) \\
 & \Rightarrow 2{{\tan }^{-1}}\dfrac{1}{\sqrt{2}-1}={{\tan }^{-1}}\left( -1 \right) \\
\end{align}\]
From inverse trigonometric functions, we have ${{\tan }^{-1}}\left( -1 \right)=\dfrac{3\pi }{4}$. Hence, we can say from the above equation that \[2{{\tan }^{-1}}\dfrac{1}{\sqrt{2}-1}=\dfrac{3\pi }{4}\]. Since we had simplified $2{{\cot }^{-1}}\left( \sqrt{2}-1 \right)$ to \[2{{\tan }^{-1}}\dfrac{1}{\sqrt{2}-1}\], so finally, we can say that \[2{{\cot }^{-1}}\left( \sqrt{2}-1 \right)=\dfrac{3\pi }{4}\].
Since we got \[2{{\cot }^{-1}}\left( \sqrt{2}-1 \right)=\dfrac{3\pi }{4}\], substituting this in the expression given in the question i.e. ${{\cos }^{-1}}\left\{ \cos 2{{\cot }^{-1}}\left( \sqrt{2}-1 \right) \right\}$, we get ${{\cos }^{-1}}\left\{ \cos \left( \dfrac{3\pi }{4} \right) \right\}$ .
The angle inside the ${{\cos }^{-1}}\cos $ function is a primary angle since it is less that $\pi $ and greater than $0$. So, we can apply formula $\left( 1 \right)$ to ${{\cos }^{-1}}\left\{ \cos \left( \dfrac{3\pi }{4} \right) \right\}$.
Using formula $\left( 1 \right)$, we get ${{\cos }^{-1}}\left\{ \cos \left( \dfrac{3\pi }{4} \right) \right\}=\dfrac{3\pi }{4}$.
Hence, the answer is option (c).

Note: One must know that the formula ${{\cos }^{-1}}\left( \cos x \right)=x$is valid only when $x$ is a primary angle i.e. $x\in \left[ 0,\pi \right]$. One cannot use this formula if $x$ is not a primary angle i.e. $x\notin \left[ 0,\pi \right]$.