Question:

Assertion (A): The principal value of $\cot^{-1}(\sqrt{3})$ is $\frac{\pi}{6}$.
Reason (R): Domain of $\cot^{-1} x$ is $\mathbb{R} - \{-1, 1\}$.

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The domain of inverse cotangent is $\mathbb{R}$; range is $(0, \pi)$ for principal value.
  • Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  • Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of the Assertion (A)
  • Assertion (A) is true, but Reason (R) is false.
  • Assertion (A) is false, but Reason (R) is true.
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The Correct Option is C

Solution and Explanation

First, check the Assertion: \[ \cot^{-1}(\sqrt{3}) = \theta. \text{So,} \cot \theta = \sqrt{3}. \] We know, \[ \cot \frac{\pi}{6} = \frac{\cos \frac{\pi}{6}}{\sin \frac{\pi}{6}} = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}. \] Also, the principal value branch of $\cot^{-1} x$ is $(0, \pi)$, so the angle $\frac{\pi}{6}$ is valid and in principal range. So, the Assertion is True. Now, check the Reason: The domain of the inverse cotangent function is: \[ \text{Domain}(\cot^{-1} x) = \mathbb{R}, \text{because for any real number } x, \cot^{-1} x \text{ is defined.} \] There is no restriction like $\{-1, 1\}$. Maybe the confusion is with $\tan^{-1} x$ where domain is also $\mathbb{R}$. So the Reason is False. Therefore, Assertion (A) is true but Reason (R) is false.
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