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Mark the tick against the correct answer in the following: The principal value of \cot ^{-1}(-\sqrt{3}) is
A. \frac{2 \pi}{6}
B. \frac{\pi}{6}
C. \frac{7 \pi}{6}
D. \frac{5 \pi}{6}
CBSE | Class 12 | Inverse Trigonometric Functions | Maths | RS Aggarwal
Mark the tick against the correct answer in the following: The principal value of \cot ^{-1}(-\sqrt{3}) is
A. \frac{2 \pi}{6}
B. \frac{\pi}{6}
C. \frac{7 \pi}{6}
D. \frac{5 \pi}{6}

Solution:

Option(D) is correct.
To Find: The Principle value of \cot ^{-1}(-\sqrt{3})
Let the principle value be given by \mathrm{x}
Now, let x=\cot ^{-1}(-\sqrt{3})
\begin{array}{l} \Rightarrow \cot x=-\sqrt{3} \\ \Rightarrow \cot x=-\cot \left(\frac{\pi}{6}\right)\left(\because \cot \left(\frac{\pi}{6}\right)=\sqrt{3}\right) \\ \Rightarrow \cot x=\cot \left(\pi-\frac{\pi}{6}\right)(\because-\cot (\theta)=\cot (\pi-\theta)) \\ \Rightarrow x=\frac{5 \pi}{6} \end{array}

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