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

Find the principal value of cot1(13){{\cot }^{-1}}\left( -\dfrac{1}{\sqrt{3}} \right).

Explanation

Solution

Hint: Find the value of angle for which its cotangent is (13)\left( -\dfrac{1}{\sqrt{3}} \right), in the range of angle (0,π)\left( 0,\pi \right). Assume this angle as θ\theta and write the above expression as cot1(cotθ){{\cot }^{-1}}\left( \cot \theta \right). Now, simply remove the function cot1{{\cot }^{-1}} and cot\cot and write the value of θ\theta as the principal value.

Complete step-by-step answer:

Since, none of the six trigonometric functions are one-to-one, they are restricted in order to have inverse functions. Therefore, the ranges of the inverse functions are proper subsets of the domains of the original functions.
Now let us come to the question. We have to find the principal value of cot1(13){{\cot }^{-1}}\left( -\dfrac{1}{\sqrt{3}} \right).
We know that the range of cot1x{{\cot }^{-1}}x is between 0 and π\pi . So, we have to select such a value of the angle that must lie between 0 and π\pi and its cotangent is 13\dfrac{-1}{\sqrt{3}}.
We know that the value of cotangent is (13)\left( \dfrac{-1}{\sqrt{3}} \right) when the angle is 2π3\dfrac{2\pi }{3}, which lies between 0 and π\pi . Clearly we can see that this angle lies in the 2nd quadrant and therefore its cotangent is negative. Therefore, the expression cot1(13){{\cot }^{-1}}\left( -\dfrac{1}{\sqrt{3}} \right) can be written as:
cot1(13)=cot1(cot2π3){{\cot }^{-1}}\left( \dfrac{-1}{\sqrt{3}} \right)={{\cot }^{-1}}\left( \cot \dfrac{2\pi }{3} \right)
We know that,
cot1(cotx)=x{{\cot }^{-1}}\left( \cot x \right)=x, when ‘x’ lies between 0 and π\pi .
Since, 2π3\dfrac{2\pi }{3} lies between 0 and π\pi . Therefore,
cot1(cot2π3)=2π3{{\cot }^{-1}}\left( \cot \dfrac{2\pi }{3} \right)=\dfrac{2\pi }{3}
Hence, the principal value of cot1(13){{\cot }^{-1}}\left( -\dfrac{1}{\sqrt{3}} \right) is 2π3\dfrac{2\pi }{3}.

Note: One may note that there is only one principal value of an inverse trigonometric function. We know that at many angles the value of cot is (13)\left( -\dfrac{1}{\sqrt{3}} \right) but we have to remember the range in which cot inverse function is defined. We have to choose such an angle which lies in the range and satisfies the function. So, there can be only one answer.