Question

Prove that $ \cot \left( {{\cot }^{-1}}x \right)=x,x\in \mathbb{R} $

Answer 1

Hint: Use the fact that if $ y={{\cot }^{-1}}x $ , then $ x=\cot y $ . Assume $ y={{\cot }^{-1}}x $ . Write $ \cot \left( {{\cot }^{-1}}x \right) $ in terms of y and hence prove the above result.

Complete step-by-step answer:
Before dwelling into the proof of the above question, we must understand how $ {{\cot }^{-1}}x $ is defined even when $ \cot x $ is not one-one.
We know that cotx is a periodic function.
Let us draw the graph of cotx

As is evident from the graph cotx is a repeated chunk of the graph of cotx within the interval (A, B) and it attains all its possible values in the interval (A, B)
Hence if we consider cotx in the interval (A, B), we will lose no value attained by cotx, and at the same time, cotx will be one-one and onto.
Hence $ {{\cot }^{-1}}x $ is defined over the Domain $ \mathbb{R} $ , with codomain $ \left( 0,\pi \right) $ as in the Domain $ \left( 0,\pi \right) $ , cotx is one-one and $ \text{Range}\left( \cot x \right)=\mathbb{R} $ .
Now since $ {{\cot }^{-1}}x $ is the inverse of cotx it satisfies the fact that if $ y={{\cot }^{-1}}x $ , then $ \cot y=x $ .
So let $ y={{\cot }^{-1}}x $ .
Hence we have coty = x.
Now $ \cot \left( {{\cot }^{-1}}x \right)=\cot y $
Hence we have $ \cot \left( {{\cot }^{-1}}x \right)=x $ .
Also as x is the Domain of $ {{\cot }^{-1}}x $ , we have $ x\in \mathbb{R} $ .
Hence $ \cot \left( {{\cot }^{-1}}x \right)=x,x\in \mathbb{R} $

Note: [1] The above-specified codomain for $ {{\cot }^{-1}}x $ is called principal branch for $ {{\cot }^{-1}}x $ . We can select any branch as long as $ \cot x $ is one-one and onto and Range $ =\mathbb{R} $ . Instead of $ \left( 0,\pi \right) $ , we can select the interval $ \left( \pi ,2\pi \right) $ . The proof will remain the same as above.