Find the domain of $f\left( x \right)=\dfrac{1}{\sqrt{\ln \left( {{\cot }^{-1}}x \right)}}$?
(a) $\left( \cot 1,\infty \right)$
(b) $R-\cot 1$
(c) $\left( -\infty ,0 \right)\cup \left( 0,\cot 1 \right)$
(d) $\left( -\infty ,\cot 1 \right)$

Hint: To find the domain of the function, find the set of all possible values of x for which the function is defined. For the log function to be defined consider its argument greater than 0 and for the square root in the denominator to be defined consider the term inside the square root greater than 0. Use the fact that the inverse cotangent function is a decreasing function to form the set of values of x.

Complete step by step answer:
Here we have been provided with the function $f\left( x \right)=\dfrac{1}{\sqrt{\ln \left( {{\cot }^{-1}}x \right)}}$ and we are asked to find the domain. First we need to understand the term domain.
Now, in mathematics the term domain is the set of values of x for which the function is defined. As we can see that we have a logarithmic term inside the square root symbol and which is present in the denominator. For a function to be defined its denominator must not be 0 and the term inside the square root must be greater than or equal to 0.
Since the function $\ln \left( {{\cot }^{-1}}x \right)$ is in the denominator and also inside the square root so it must not be 0 or less than 0, so we have,
$\Rightarrow \ln \left( {{\cot }^{-1}}x \right)>0$
We can 0 as $\ln 1$ and as the base of log on both the sides is e (natural log) so when we will remove the log function the direction of the inequality will remain same, so we get,
\[\begin{align}
  & \Rightarrow \ln \left( {{\cot }^{-1}}x \right)>\ln 1 \\
 & \Rightarrow {{\cot }^{-1}}x>1 \\
\end{align}\]
1 in the R.H.S means 1 radian which is in the range $\left( 0,\pi \right)$ that is the range of ${{\cot }^{-1}}x$ so we can write $1={{\cot }^{-1}}\left( \cot x \right)$, therefore we have,
\[\Rightarrow {{\cot }^{-1}}x>{{\cot }^{-1}}\left( \cot 1 \right)\]
We know that the inverse cotangent function is a decreasing function so when we will remove that function from both the sides then the direction of the inequality will get reversed, so we get,
\[\Rightarrow x<\cot 1\] ……… (1)
Now, the argument of the log function must be greater than 0 to be defined, so we get,
$\Rightarrow {{\cot }^{-1}}x>0$
We know that the range of ${{\cot }^{-1}}x$ is $\left( 0,\pi \right)$ which is greater than 0 for all real values of x, so we get,
$\Rightarrow -\infty < x <\infty $ ……… (2)
We have to take those values of x such that both the relations (1) and (2) gets satisfied that means we have to take the intersection of the two sets, so we get,
$\begin{align}
  & \Rightarrow x\in \left( -\infty ,\infty \right)\cap \left( -\infty ,\cot 1 \right) \\
 & \therefore x\in \left( -\infty ,\cot 1 \right) \\
\end{align}$

So, the correct answer is “Option d”.

Note: You must remember the range and domain values of all the trigonometric, inverse trigonometric, logarithmic, exponential functions. Note that if the base value of log is between 0 and 1 then the direction of inequality gets changed while removing the function. Remember the increasing and decreasing nature of common functions which determine the change of inequality sign while removing them.