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

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

Explanation

Solution

We will first assume that cot⁡−1x=y{\cot ^{ - 1}}x = y. Then, we can write it as: x=cot⁡y=1tan⁡yx = \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⁡−1x=tan⁡−1(1x){\cot ^{ - 1}}x = {\tan ^{ - 1}}\left( {\dfrac{1}{x}} \right) for x>0x > 0.
Since we know that the principal argument for cotangent of an angle is [0,π]\left[ {0,\pi } \right] and here, we have x>0x > 0, therefore, the condition is x∈(0,π2)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:-
⇒cot⁡−1x=y\Rightarrow {\cot ^{ - 1}}x = y ………………………(1)
Taking cot on both the sides of above equation, we will then obtain the following equation with us:-
⇒cot⁡(cot⁡−1x)=cot⁡y\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:-
⇒x=cot⁡y\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=1tan⁡y\cot y = \dfrac{1}{{\tan y}}.
Putting this in the last mentioned equation, we will then obtain the following equation with us:-
⇒x=1tan⁡y\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:-
⇒1x=11tan⁡y\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:-
⇒1x=11×tan⁡y\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:-
⇒1x=tan⁡y\Rightarrow \dfrac{1}{x} = \tan y
Taking tan⁡−1{\tan ^{ - 1}} on both the sides of the above equation, we will then obtain the following equation with us:-
⇒tan⁡−1(1x)=tan⁡−1(tan⁡y)\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:-
⇒tan⁡−1(1x)=y\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:-
⇒cot⁡−1x=tan⁡−1(1x)\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∈[π2,π)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.