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Question: Write the value of \({{\cot }^{-1}}\left( -x \right)\) for all \(x\in R\) in terms of \({{\cot }^{-1...

Write the value of cot⁡−1(−x){{\cot }^{-1}}\left( -x \right) for all x∈Rx\in R in terms of cot⁡−1x{{\cot }^{-1}}x.

Explanation

Solution

Hint: In this given question, we must first of all convert the given equation, that is cot⁡−1(−x){{\cot }^{-1}}\left( -x \right) in terms of cot⁡\cot . Thereafter we can use the property of cotangent function to express x as a cotangent of some angle. Then, we can again take cot⁡−1{{\cot }^{-1}} on both sides and use the given relation to obtain the required answer.

Complete step-by-step answer:

In this given question, we are first of all going to convert the given equation, that is cot⁡−1(−x){{\cot }^{-1}}\left( -x \right) in terms of cot⁡\cot as follows:
Let, cot⁡−1(−x)=θ{{\cot }^{-1}}\left( -x \right)=\theta ………………….(1.1)
⇒(−x)=cot⁡θ\Rightarrow \left( -x \right)=\cot \theta
⇒x=−cot⁡θ........................(1.2)\Rightarrow x=-\cot \theta ........................(1.2)
Now, we know that when we increase the value of an angle by a multiple of π\pi , the value of its cotangent becomes negative, thus we can write
cot⁡(nπ+θ)=−cot⁡(θ)...............(1.3)\cot (n\pi +\theta )=-\cot \left( \theta \right)...............(1.3)
where n is an integer.
Using this in equation (1.2), we get
x=cot⁡(nπ−θ)x=\cot \left( n\pi -\theta \right)
⇒cot⁡−1(x)=nπ−θ\Rightarrow {{\cot }^{-1}}\left( x \right)=n\pi -\theta ………………………. (1.4)
Now, from equation 1.1, we get the value of θ\theta as equal to cot⁡−1(−x){{\cot }^{-1}}\left( -x \right). So, putting this value in equation 1.4, we get,
cot⁡−1(x)=nπ−θ{{\cot }^{-1}}\left( x \right)=n\pi -\theta
⇒cot⁡−1(x)=nπ−cot⁡−1(−x)\Rightarrow {{\cot }^{-1}}\left( x \right)=n\pi -{{\cot }^{-1}}\left( -x \right)
⇒cot⁡−1(−x)=nπ−cot⁡−1(x)\Rightarrow {{\cot }^{-1}}\left( -x \right)=n\pi -{{\cot }^{-1}}\left( x \right)……………… (1.5)
Hence, from equation 1.5, we get the required expression we wanted to prove.
Therefore, we have expressed the value of cot⁡−1(−x){{\cot }^{-1}}\left( -x \right) for all x∈Rx\in R in terms of cot⁡−1x{{\cot }^{-1}}x.

Note: We should note that in equation (1.2), we could have written a more general expression for x as
x=−cot⁡(θ+rπ)x=-\cot \left( \theta +r\pi \right) where r in an integer because the value of cot remains the same if we increase the value of the angle by a multiple of π\pi . However, in this case, we would have got equation (1.4) as cot⁡−1(x)=nπ−(θ+rπ){{\cot }^{-1}}\left( x \right)=n\pi -\left( \theta +r\pi \right) which is equivalent to nπ−cot⁡−1(x)n\pi -{{\cot }^{-1}}\left( x \right). Thus, in both the methods, the final answer will remain the same.