Question

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

Answer 1

To solve this question, we should have some knowledge about the range of principal value of ${{\cot }^{-1}}x=\theta$, that is $0<\theta <\pi$. Here, we will first try to write $\sqrt{3}$ in terms of $\cot \theta$, and 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.
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Complete step-by-step answer:
In this question, we are 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 the equation as, $30{}^\circ ={{\cot }^{-1}}\sqrt{3}$.
So, we can say that $30{}^\circ$ can be the principal value of ${{\cot }^{-1}}\left( \sqrt{3} \right)$. But we have to check the answer, so we will try to find out whether $30{}^\circ$ lies in the range of the principal value of ${{\cot }^{-1}}x$, that is $0<\theta <\pi$. For doing that, we have to convert $30{}^\circ$ in terms of radian. Now, we know that, $180{}^\circ =\pi \text{ }radian$, which can also be written as $1{}^\circ =\dfrac{\pi }{180}\text{ }radian$. So, we can write $30{}^\circ =30\times \dfrac{\pi }{180}\text{=}\dfrac{\pi }{6}\text{ }radian$.
And we know that general solution of $\theta ={{\cot }^{-1}}x$, is $\theta =n\left( \pi \right)+{{\cot }^{-1}}x$. Therefore, we can say that the general solution of $\theta =n\left( \pi \right)+\dfrac{\pi }{6}$. So, to get the principal, we will put the value of n as 0. Therefore, we get $\theta =\dfrac{\pi }{6}$. We know that $\dfrac{\pi }{6}$ is always less than $\pi$ and greater than 0, so we can write $0<\dfrac{\pi }{6}<\pi$.
Hence, we can say that $30{}^\circ$ is the principal value of ${{\cot }^{-1}}\left( \sqrt{3} \right)$.

Note: In this question, one can make a mistake if they do not know the cotangent ratios like $\cot 30{}^\circ =\sqrt{3}$. We also need to know that the range of ${{\cot }^{-1}}x=\theta$ is $0<\theta <\pi$. And therefore, if we do not find the value of the range, then we have to transform our value to this range.