Question

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

Answer 1

Hint: In order to solve this question, we should know that the range of the principal value of ${{\cot }^{-1}}x=\theta $ is $0<\theta <\pi $. So, here we will try to convert $-\dfrac{1}{\sqrt{3}}$ of ${{\cot }^{-1}}\left( -\dfrac{1}{\sqrt{3}} \right)$ in terms of $\cot \theta $ and then we then we will find the general solution to get principal solution by using a rule, if $\theta ={{\cot }^{-1}}x$, then $\theta =n\left( \pi \right)+{{\cot }^{-1}}x$. Therefore, we will get our answer which will lie in range as principal value. Also, we need to know that $\cot 60{}^\circ =\dfrac{1}{\sqrt{3}}$.

Complete step-by-step answer:
In this question, we have been asked to find the principal value of ${{\cot }^{-1}}\left( -\dfrac{1}{\sqrt{3}} \right)$. For finding that, we should have some knowledge about the cotangent ratios like $\cot 60{}^\circ =\dfrac{1}{\sqrt{3}}$. Now we know that, $\dfrac{1}{\sqrt{3}}=\cot 60{}^\circ $. So, we can say that $-\dfrac{1}{\sqrt{3}}=-\cot 60{}^\circ $. Now, we also know that $\cot \left( 180{}^\circ -\theta \right)=-\cot \theta $. So, we can write $-\cot 60{}^\circ $ as $\cot \left( 180{}^\circ -60{}^\circ \right)$, which is nothing but $\cot 120{}^\circ $. So, we get, $\cot 120{}^\circ =-\dfrac{1}{\sqrt{3}}$. And we know that it can be further written as, $120{}^\circ ={{\cot }^{-1}}\left( -\dfrac{1}{\sqrt{3}} \right)$.
Now, if we talk about the range, that is whether $120{}^\circ $ lies in the range of the principal value of ${{\cot }^{-1}}x=\theta $, that is $0<\theta <\pi $or not, we should know that $\pi \text{ }radian=180{}^\circ $. So, the range can be written as, $0<\theta <180{}^\circ $.
In the given question, we have got the value of ${{\cot }^{-1}}\left( -\dfrac{1}{\sqrt{3}} \right)$ as $120{}^\circ $. And we know that the general solution of $\theta ={{\cot }^{-1}}x$, is $\theta =n\left( 180{}^\circ \right)+{{\cot }^{-1}}x$. Therefore, we can say that the general solution of $\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 $\theta =120{}^\circ $. And, we know that $0{}^\circ <120{}^\circ <180{}^\circ $, so it lies in the range.
Hence, we can say that, $120{}^\circ $ is the principal value of ${{\cot }^{-1}}\left( -\dfrac{1}{\sqrt{3}} \right)$.

Note: We should remember that $\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 $\cot 60{}^\circ =\dfrac{1}{\sqrt{3}}$. In this question, there is a chance of mistake if we ignore the negative sign with $\dfrac{1}{\sqrt{3}}$, which will lead to a wrong answer.