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

Find the principal value of tan1(13){{\tan}^{-1}}\left( -\dfrac{1}{\sqrt{3}} \right).

Explanation

Solution

Find the value of angle for which its tangent 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 tan1(tanθ){{\tan }^{-1}}\left( \tan \theta \right). Now, simply remove the function tan1{{\tan}^{-1}} and tan\tan and write the value of θ\theta as the principal value.

Complete step-by-step solution:
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 tan1(13){{\tan }^{-1}}\left( -\dfrac{1}{\sqrt{3}} \right).
We know that the range of tan1x{{\tan }^{-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 tangent is 13\dfrac{-1}{\sqrt{3}}.
We know that, the value of tangent is (13)\left( \dfrac{-1}{\sqrt{3}} \right) when the angle is 5π6\dfrac{5\pi }{6}, which lies between 0 and π\pi . Clearly we can see that this angle lies in the 2nd quadrant and therefore its tangent is negative. Therefore, the expression tan1(13){{\tan}^{-1}}\left( -\dfrac{1}{\sqrt{3}} \right) can be written as:
tan1(13)=tan1(tan5π6){{\tan }^{-1}}\left( \dfrac{-1}{\sqrt{3}} \right)={{\tan }^{-1}}\left( \tan \dfrac{5\pi }{6} \right)
We know that,
tan1(tanx)=x{{\tan }^{-1}}\left( \tan x \right)=x, when ‘x’ lies between 0 and π\pi .
Since, 5π6\dfrac{5\pi }{6} lies between 0 and π\pi . Therefore,
tan1(tan5π6)=5π6{{\tan }^{-1}}\left( \tan \dfrac{5\pi }{6} \right)=\dfrac{5\pi }{6}
Hence, the principal value of tan1(13){{\tan }^{-1}}\left( -\dfrac{1}{\sqrt{3}} \right) is 5π6\dfrac{5\pi }{6}.

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 tan 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.