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Question

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A.$\dfrac{\pi }{4}$

B.\[ - \dfrac{\pi }{4}\]

C.$\dfrac{{3\pi }}{4}$

D.None of these

Answer

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We have to find the principal value of ${\cot ^{ - 1}}\left( { - 1} \right)$. That is, we want to find the angle where $\cot x = - 1$ in the cartesian plane such that $0 \leqslant x < 2\pi $

We will know let ${\cot ^{ - 1}}\left( { - 1} \right) = x$, which implies that $\cot x = - 1$

Since, the value is negative, and cot of any angle is negative in the second and fourth quadrant only.

Hence, the value of $x$ is in the second and fourth quadrant.

We also know that \[\cot \left( {\dfrac{\pi }{4}} \right) = 1\]

Therefore, we will have the corresponding angle in the second and fourth quadrant.

Let us first find the value in the second quadrant.

For, writing the angle in second quadrant whose value will be negative of \[\cot \left( {\dfrac{\pi }{4}} \right)\] , we will subtract $\dfrac{\pi }{4}$ from $\pi $.

Hence, we have

$

\cot \left( {\pi - \dfrac{\pi }{4}} \right) = - 1 \\

\cot \left( {\dfrac{{3\pi }}{4}} \right) = - 1 \\

$

Hence, the principal value is \[\dfrac{{3\pi }}{4}\] in the second quadrant.

Now, we will find the value in the fourth quadrant.

For, writing the angle in second quadrant whose value will be negative of \[\cot \left( {\dfrac{\pi }{4}} \right)\] , we will subtract $\dfrac{\pi }{4}$ from $2\pi $.

Hence, we have

$

\cot \left( {2\pi - \dfrac{\pi }{4}} \right) = - 1 \\

\Rightarrow \cot \left( {\dfrac{{7\pi }}{4}} \right) = - 1 \\

$

Hence, the principal value is \[\dfrac{{7\pi }}{4}\] in the fourth quadrant.