Question

The value of the trigonometric function $\tan \left( {{\cot }^{-1}}x \right)$ is equal to:
\[\begin{align}
  & A.\dfrac{\pi }{2}-x \\
 & B.\cot \left( {{\tan }^{-1}}x \right) \\
 & C.\tan x \\
 & D.\dfrac{1}{x} \\
\end{align}\]

Answer 1

Hint: The above question is based on the properties of Inverse circular function. Some properties which are used in our question is shown below:
\[\begin{align}
  & \left( i \right)y=\tan \left( {{\tan }^{-1}}x \right)=x,x\in R,y\in R,\text{ y is a periodic} \\
 & \left( ii \right)y=\cot \left( {{\cot }^{-1}}x \right)=x,x\in R,y\in R,\text{ y is a periodic} \\
 & \left( iii \right){{\cot }^{-1}}x={{\tan }^{-1}}\dfrac{1}{x};x\text{ }>\text{ 0} \\
 & \Rightarrow \pi +{{\tan }^{-1}}\dfrac{1}{x};x\text{ }<\text{ 0} \\
 & \left( iv \right){{\tan }^{-1}}x+{{\cot }^{-1}}x=\dfrac{\pi }{2};x\in R \\
\end{align}\]
These are basic identities or properties of Inverse trigonometric function. We have to apply them to our question. There may be more than one solution.

Complete step by step answer:
Let's move to our problem, we have $\tan \left( {{\cot }^{-1}}x \right)$ expression.
Let’s take it as \[P=\tan \left( {{\cot }^{-1}}x \right)\cdots \cdots \cdots \cdots \left( i \right)\]
Now, by property \[{{\tan }^{-1}}x+{{\cot }^{-1}}x=\dfrac{\pi }{2};x\in R\]
We have \[{{\cot }^{-1}}x=\dfrac{\pi }{2}-{{\tan }^{-1}}x\]
Now, put this value in (i).
Hence, we have \[P=\tan \left( \dfrac{\pi }{2}-{{\tan }^{-1}}x \right)\]
We know that \[\tan \left( \dfrac{\pi }{2}-\theta \right)=\cot \theta \]
Hence, \[P=\cot \left( {{\tan }^{-1}}x \right)\cdots \cdots \cdots \cdots \left( ii \right)\]
So, this is one of our answer according to options given in the question i.e. option B.
Now, by property
\[\begin{align}
  & {{\cot }^{-1}}x={{\tan }^{-1}}\dfrac{1}{x};x\text{ }>\text{ 0} \\
 & \Rightarrow \pi +{{\tan }^{-1}}\dfrac{1}{x};x\text{ }<\text{ 0} \\
\end{align}\]
We have from (i).
\[\begin{align}
  & P=\tan \left( {{\cot }^{-1}}x \right) \\
 & P=\tan \left( {{\tan }^{-1}}\left( \dfrac{1}{x} \right) \right)\left( \text{when x0} \right) \\
\end{align}\]
Now, we know, property \[\tan \left( {{\tan }^{-1}}x \right)=x,x\in R\]
Hence, we get
\[\begin{align}
  & P=\tan \left( {{\tan }^{-1}}\left( \dfrac{1}{x} \right) \right) \\
 & P=\left( \dfrac{1}{x} \right) \\
\end{align}\]
Now, we have one other possibility when x<0
\[\begin{align}
  & \therefore P=\tan \left( {{\cot }^{-1}}x \right) \\
 & P=\tan \left( \pi +{{\tan }^{-1}}\left( \dfrac{1}{x} \right) \right) \\
\end{align}\]
Now, we know \[\tan \left( \pi +\theta \right)=\tan \theta \]
Hence, \[P=\tan \left( \pi +{{\tan }^{-1}}\left( \dfrac{1}{x} \right) \right)=\tan \left( {{\tan }^{-1}}\left( \dfrac{1}{x} \right) \right)=\left( \dfrac{1}{x} \right)\]
Hence, in both cases i.e. x>0 and x<0
We get \[P=\dfrac{1}{x}\]
So, this is our second answer i.e. option D.
Hence, overall we have options B and D as our final answer.

Note: Inverse trigonometric function becomes quite easy to solve, once we grab the formula and properties. Students can make mistake in understanding the difference between \[\tan \left( {{\tan }^{-1}}x \right)\text{ and }{{\tan }^{-1}}\left( \tan x \right)\]. Both terms are looking similar but they are quite different in their formula like:
\[\tan \left( {{\tan }^{-1}}x \right)=x,x\in R\]
\[{{\tan }^{-1}}\left( \tan x \right)=\left\{ \begin{align}
  & x+\pi ,\dfrac{-3\pi }{2}\text{ }<\text{ }x\text{ }<\text{ }\dfrac{-\pi }{2} \\
 & x,\dfrac{-\pi }{2}\text{ }<\text{ }x\text{ }<\text{ }\dfrac{\pi }{2} \\
 & x-\pi ,\dfrac{\pi }{2}\text{ }<\text{ }x\text{ }<\text{ }\dfrac{3\pi }{2} \\
 & x-2\pi ,\dfrac{3\pi }{2}\text{ }<\text{ }x\text{ }<\text{ }\dfrac{5\pi }{2} \\
 & x-3\pi ,\dfrac{5\pi }{2}\text{ }<\text{ }x\text{ }<\text{ }\dfrac{7\pi }{2} \\
\end{align} \right.\]