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Question: Find the principal value of \({{\cot }^{-1}}\left( -\dfrac{1}{\sqrt{3}} \right)\)....

Find the principal value of cot⁡−1(−13){{\cot }^{-1}}\left( -\dfrac{1}{\sqrt{3}} \right).

Explanation

Solution

In order to solve this question, we should know that the range of the principal value of cot⁡−1x=θ{{\cot }^{-1}}x=\theta is 0<θ<π0<\theta <\pi . So, here we will try to convert −13-\dfrac{1}{\sqrt{3}} of cot⁡−1(−13){{\cot }^{-1}}\left( -\dfrac{1}{\sqrt{3}} \right) in terms of cot⁡θ\cot \theta and then we then we will find the general solution to get principal solution by using a rule, if θ=cot⁡−1x\theta ={{\cot }^{-1}}x, then θ=n(π)+cot⁡−1x\theta =n\left( \pi \right)+{{\cot }^{-1}}x. Therefore, we will get our answer which will lie in range as principal value. Also, we need to know that cot⁡60∘=13\cot 60{}^\circ =\dfrac{1}{\sqrt{3}}.
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Complete step-by-step answer:
In this question, we have been asked to find the principal value of cot⁡−1(−13){{\cot }^{-1}}\left( -\dfrac{1}{\sqrt{3}} \right). For finding that, we should have some knowledge about the cotangent ratios like cot⁡60∘=13\cot 60{}^\circ =\dfrac{1}{\sqrt{3}}. Now we know that, 13=cot⁡60∘\dfrac{1}{\sqrt{3}}=\cot 60{}^\circ . So, we can say that −13=−cot⁡60∘-\dfrac{1}{\sqrt{3}}=-\cot 60{}^\circ . Now, we also know that cot⁡(180∘−θ)=−cot⁡θ\cot \left( 180{}^\circ -\theta \right)=-\cot \theta . So, we can write −cot⁡60∘-\cot 60{}^\circ as cot⁡(180∘−60∘)\cot \left( 180{}^\circ -60{}^\circ \right), which is nothing but cot⁡120∘\cot 120{}^\circ . So, we get, cot⁡120∘=−13\cot 120{}^\circ =-\dfrac{1}{\sqrt{3}}. And we know that it can be further written as, 120∘=cot⁡−1(−13)120{}^\circ ={{\cot }^{-1}}\left( -\dfrac{1}{\sqrt{3}} \right).
Now, if we talk about the range, that is whether 120∘120{}^\circ lies in the range of the principal value of cot⁡−1x=θ{{\cot }^{-1}}x=\theta , that is 0<θ<π0<\theta <\pi or not, we should know that π radian=180∘\pi \text{ }radian=180{}^\circ . So, the range can be written as, 0<θ<180∘0<\theta <180{}^\circ .
In the given question, we have got the value of cot⁡−1(−13){{\cot }^{-1}}\left( -\dfrac{1}{\sqrt{3}} \right) as 120∘120{}^\circ . And we know that the general solution of θ=cot⁡−1x\theta ={{\cot }^{-1}}x, is θ=n(180∘)+cot⁡−1x\theta =n\left( 180{}^\circ \right)+{{\cot }^{-1}}x. Therefore, we can say that the general solution of θ=n(180∘)+120∘\theta =n\left( 180{}^\circ \right)+120{}^\circ . So, to get the principal, we will put the value of n as 0. Therefore, we get θ=120∘\theta =120{}^\circ . And, we know that 0∘<120∘<180∘0{}^\circ <120{}^\circ <180{}^\circ , so it lies in the range.
Hence, we can say that, 120∘120{}^\circ is the principal value of cot⁡−1(−13){{\cot }^{-1}}\left( -\dfrac{1}{\sqrt{3}} \right).

Note: We should remember that cot⁡(π−θ)=−cot⁡θ\cot \left( \pi -\theta \right)=-\cot \theta , while solving this question, because we will require this property while finding the principal value. Also, we should know that cot⁡60∘=13\cot 60{}^\circ =\dfrac{1}{\sqrt{3}}. In this question, there is a chance of mistake if we ignore the negative sign with 13\dfrac{1}{\sqrt{3}}, which will lead to a wrong answer.