Question

How do you prove ${\cot ^{ - 1}}x = {\tan ^{ - 1}}\left( {\dfrac{1}{x}} \right)$ for $x > 0$?

Answer 1

Hint: We will first assume that ${\cot ^{ - 1}}x = y$. Then, we can write it as: $x = \cot y = \dfrac{1}{{\tan y}}$.
Thus, now we can invert both the sides and take inverse of tangent to get the required answer.

Complete step by step answer:
We are given that we are required to prove ${\cot ^{ - 1}}x = {\tan ^{ - 1}}\left( {\dfrac{1}{x}} \right)$ for $x > 0$.
Since we know that the principal argument for cotangent of an angle is $\left[ {0,\pi } \right]$ and here, we have $x > 0$, therefore, the condition is $x \in \left( {0,\dfrac{\pi }{2}} \right)$.
Now, let us assume the left hand side of the above equation to be equal to y, we will then obtain the following equation:-
$ \Rightarrow {\cot ^{ - 1}}x = y$ ………………………(1)
Taking cot on both the sides of above equation, we will then obtain the following equation with us:-
$ \Rightarrow \cot \left( {{{\cot }^{ - 1}}x} \right) = \cot y$
Simplifying the left hand side of the above equation, we will then obtain the following equation with us:-
$ \Rightarrow x = \cot y$
Now, since we know that tangent of an angle is inverse of cotangent of that angle.
So, we have: $\cot y = \dfrac{1}{{\tan y}}$.
Putting this in the last mentioned equation, we will then obtain the following equation with us:-
$ \Rightarrow x = \dfrac{1}{{\tan y}}$
Taking reciprocal of both the sides of the above equation, we will then obtain the following equation with us:-
$ \Rightarrow \dfrac{1}{x} = \dfrac{1}{{\dfrac{1}{{\tan y}}}}$
Simplifying the right hand side of the above equation, we will then obtain the following equation with us:-
$ \Rightarrow \dfrac{1}{x} = \dfrac{1}{1} \times \tan y$
Simplifying the right hand side of the above equation further, we will then obtain the following equation with us:-
$ \Rightarrow \dfrac{1}{x} = \tan y$
Taking ${\tan ^{ - 1}}$ on both the sides of the above equation, we will then obtain the following equation with us:-
$ \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{1}{x}} \right) = {\tan ^{ - 1}}\left( {\tan y} \right)$
Simplifying the right hand side of the above equation, we will then obtain the following equation with us:-
$ \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{1}{x}} \right) = y$ …………….(2)
Using equation number 1 and 2 together, we will then obtain the following equation as required with us:-
$ \Rightarrow {\cot ^{ - 1}}x = {\tan ^{ - 1}}\left( {\dfrac{1}{x}} \right)$

Note:
The students must note that if there would have been the possibility of x being less than or equal to 0, then the situation would have been changed and \[x \in \left[ {\dfrac{\pi }{2},\left. \pi \right)} \right.\].
Therefore, if it would have has just for any real number x, then we would have obtained the following result:-
$ \Rightarrow {\cot ^{ - 1}}x = \left\{ {\begin{array}{*{20}{c}}
  {{{\tan }^{ - 1}}\left( {\dfrac{1}{x}} \right),x > 0} \\
  {\pi + {{\tan }^{ - 1}}\left( {\dfrac{1}{x}} \right),x \leqslant 0}
\end{array}} \right.$