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Question: For the principal value, evaluate the following: \[{{\operatorname{cosec}}^{-1}}\left[ 2\tan \left...

For the principal value, evaluate the following:
cosec⁡−1[2tan⁡(11π6)]{{\operatorname{cosec}}^{-1}}\left[ 2\tan \left( \dfrac{11\pi }{6} \right) \right].

Explanation

Solution

Hint:For the above question we will have to know about the principal value of an inverse trigonometric function is a value that belongs to the principal branch of range of function. We know that the function branch of range for cosec⁡−1x{{\operatorname{cosec}}^{-1}}x is \left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]-\left\\{ 0 \right\\}.

Complete step-by-step answer:
We have been given to evaluate the trigonometric expression cosec⁡[2tan⁡(11π6)]\operatorname{cosec}\left[ 2\tan \left( \dfrac{11\pi }{6} \right) \right].
Now we know that the principal value means the value which lies between the defined range of inverse trigonometric function.
For cosec⁡−1x{{\operatorname{cosec}}^{-1}}x the angle is \left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]-\left\\{ 0 \right\\}.
We know that tan⁡(11π6)=tan⁡(2π−π6)=−tan⁡π6=−13\tan \left( \dfrac{11\pi }{6} \right)=\tan \left( 2\pi -\dfrac{\pi }{6} \right)=-\tan \dfrac{\pi }{6}=\dfrac{-1}{\sqrt{3}}.
Since tan⁡(2π−θ)=−tan⁡θ\tan \left( 2\pi -\theta \right)=-\tan \theta as the value lies in the fourth quadrant and tangent value is negative in that quadrant
On substituting the values of tan⁡11π6\tan \dfrac{11\pi }{6} in the given expression, we get as follows:
cosec⁡−1(2tan⁡11π6)=cosec⁡−1[2(−13)]=cosec⁡−1(−23){{\operatorname{cosec}}^{-1}}\left( 2\tan \dfrac{11\pi }{6} \right)={{\operatorname{cosec}}^{-1}}\left[ 2\left( \dfrac{-1}{\sqrt{3}} \right) \right]={{\operatorname{cosec}}^{-1}}\left( \dfrac{-2}{\sqrt{3}} \right)
We know that cosec⁡(−π3)=−23\operatorname{cosec}\left( \dfrac{-\pi }{3} \right)=\dfrac{-2}{\sqrt{3}}.
By substituting the value of (−23)\left( \dfrac{-2}{\sqrt{3}} \right) in the expression, we get as follows:
cosec⁡−1(2tan⁡11π6)=cosec⁡−1[cosec⁡(−π3)]{{\operatorname{cosec}}^{-1}}\left( 2\tan \dfrac{11\pi }{6} \right)={{\operatorname{cosec}}^{-1}}\left[ \operatorname{cosec}\left( \dfrac{-\pi }{3} \right) \right]
We know that cosec⁡−1cosec⁡θ{{\operatorname{cosec}}^{-1}}\operatorname{cosec}\theta , where θ\theta must lie between the interval \left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]-\left\\{ 0 \right\\}.
⇒cosec⁡−1(2tan⁡11π6)=−π3\Rightarrow {{\operatorname{cosec}}^{-1}}\left( 2\tan \dfrac{11\pi }{6} \right)=\dfrac{-\pi }{3}
Therefore, the principal value of the given expression is equal to (−π3)\left( \dfrac{-\pi }{3} \right).

Note: Be careful while finding the principal value of the inverse trigonometric function and do check once that the value must lie between the principal branch of range of the function. Sometimes we forget the ‘2’ multiplied by tan⁡11π6\tan \dfrac{11\pi }{6} in the given expression and we just substitute the values of tan⁡11π6\tan \dfrac{11\pi }{6} and we get the incorrect answer. So be careful while solving.