Write \[{{\cot }^{-1}}\left( \dfrac{1}{\sqrt{{{x}^{2}}-1}} \right),x>1\]in the simplest form.

Hint: In this question, we first need to assume the value of x as a secant function. Then on substituting it and simplifying further by using the trigonometric identity we can write it in terms of cotangent. Now, by using properties of inverse trigonometric functions and then substituting the value of x back we get the simplest form.
\[{{\sec }^{2}}\theta -1={{\tan }^{2}}\theta \]
\[{{\cot }^{-1}}\left( \cot \theta \right)=\theta \text{ }\theta \in \left( 0,\pi \right)\]

Complete step-by-step answer:

Now, from the given expression in the question we have,
\[\Rightarrow {{\cot }^{-1}}\left( \dfrac{1}{\sqrt{{{x}^{2}}-1}} \right)\]
Now, let us assume the value of x as some secant function
\[x=\sec \theta \]
Now, on substituting the value of x in the above expression we get,
\[\Rightarrow {{\cot }^{-1}}\left( \dfrac{1}{\sqrt{{{\sec }^{2}}\theta -1}} \right)\]
As we already know from the trigonometric identity that
\[{{\sec }^{2}}\theta -1={{\tan }^{2}}\theta \]
Let us now substitute this value of trigonometric identity in the above expression to simplify it further
\[\Rightarrow {{\cot }^{-1}}\left( \dfrac{1}{\sqrt{{{\tan }^{2}}\theta }} \right)\]
Now, this can be further written as
\[\Rightarrow {{\cot }^{-1}}\left( \dfrac{1}{\tan \theta } \right)\]
As we already know that the relation between tangent function and cotangent function.
\[\cot \theta =\dfrac{1}{\tan \theta }\]
Now, by substituting this relation in the above expression obtained we get,
\[\Rightarrow {{\cot }^{-1}}\left( \cot \theta \right)\]
As we already know from the properties of inverse trigonometric functions we gte,
\[\Rightarrow \theta \]
Now, from the value of x we assumed we can further write it as
\[x=\sec \theta \]
Now, on applying inverse of secant function on both sides we get,
\[\Rightarrow \theta ={{\sec }^{-1}}x\]
Now, on substituting this value of theta back we get,
\[\Rightarrow {{\sec }^{-1}}x\]
Hence, the simplest form of \[{{\cot }^{-1}}\left( \dfrac{1}{\sqrt{{{x}^{2}}-1}} \right)\] is \[{{\sec }^{-1}}x\]

Note: It is important to note that in order to simplify the given expression we first need to assume some value of x. As the inverse function is cotangent we need to get the expression inside in terms of tangent which can be obtained by considering x as secant function.
Instead of assuming x as secant we can also assume it as cosecant and then using the trigonometric identity we can simplify it further accordingly to get the result. Both the methods give the same result.