Question

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

Answer 1

Hint: To solve this question, we should have the knowledge about the range of principal value of ${{\cot }^{-1}}x=\theta$, that is $0<\theta <\pi$. Here, we will convert $\sqrt{3}$ in terms of $\cot \theta$. And then we will find the general solution to get the principal solution by using a rule, if $\theta ={{\cot }^{-1}}x$, then $x=n\left( \pi \right)+{{\cot }^{-1}}x$. Therefore, we will get our answer which will lie in the range as principal value. Also, we need to remember that $\cot \left( \pi -\theta \right)=-\cot \theta$.
Complete step-by-step answer:
In this question, we have been asked to find the principal value of ${{\cot }^{-1}}\left( -\sqrt{3} \right)$. For that we should have some knowledge regarding the cotangent ratios like $\cot 30{}^\circ =\sqrt{3}$. Now, we know that, $\sqrt{3}=\cot 30{}^\circ$, so we can write $-\sqrt{3}=-\cot 30{}^\circ$. Now, we know that $\cot \left( 180{}^\circ -\theta \right)$ can be written as $-\cot \theta$. So, we can write, $-\sqrt{3}=\cot \left( 180{}^\circ -30{}^\circ \right)$, which is nothing but $\cot 150{}^\circ$. So, we get, $\cot 150{}^\circ =-\sqrt{3}$. And we know that it can be further written as, $150{}^\circ ={{\cot }^{-1}}\left( -\sqrt{3} \right)$.
Now, if we talk about the range, that is whether $150{}^\circ$ 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 ${{\cot }^{-1}}\left( -\sqrt{3} \right)$ as $150{}^\circ$. 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 $\theta =n\left( 180{}^\circ \right)+150{}^\circ$. So, to get the principal, we will put the value of n as 0. Therefore, we get $\theta =150{}^\circ$. And, we know that $0{}^\circ <150{}^\circ <180{}^\circ$, so it lies in the range.
Hence, we can say that, $150{}^\circ$ is the principal value of ${{\cot }^{-1}}\left( -\sqrt{3} \right)$.

Note: The possible mistake one can make while solving this question is by ignoring the negative sign with $\sqrt{3}$, because it will then lead to the incorrect answer. Also, we need to remember that if we don’t get the value in the range, then we have to apply properties like $\cot \left( \pi -\theta \right)=-\cot \theta$ to get the value in range.