Class CBSE Class 12 Mathematics Inverse Trigonometric Functions Q #586
KNOWLEDGE BASED
REMEMBER
1 Marks 2025 AISSCE(Board Exam) MCQ SINGLE
The principal value of \(\cot^{-1}(-\frac{1}{\sqrt{3}})\) is:
(A) \(-\frac{\pi}{3}\)
(B) \(-\frac{2\pi}{3}\)
(C) \(\frac{\pi}{3}\)
(D) \(\frac{2\pi}{3}\)

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Detailed Solution

Step 1: Understanding the Range of Principal Value

The principal value range of \(\cot^{-1}(x)\) is \((0, \pi)\).

Step 2: Finding the Angle

We need to find an angle \(\theta\) in the range \((0, \pi)\) such that \(\cot(\theta) = -\frac{1}{\sqrt{3}}\). Since cotangent is negative, \(\theta\) must be in the second quadrant.

Step 3: Using the Known Value

We know that \(\cot(\frac{\pi}{3}) = \frac{1}{\sqrt{3}}\). Therefore, we need to find an angle in the second quadrant with a reference angle of \(\frac{\pi}{3}\).

Step 4: Calculating the Angle in the Second Quadrant

The angle in the second quadrant is \(\pi - \frac{\pi}{3} = \frac{2\pi}{3}\). Thus, \(\cot(\frac{2\pi}{3}) = -\frac{1}{\sqrt{3}}\).

Step 5: Determining the Principal Value

Since \(\frac{2\pi}{3}\) is in the range \((0, \pi)\), the principal value of \(\cot^{-1}(-\frac{1}{\sqrt{3}})\) is \(\frac{2\pi}{3}\).

Final Answer: \(\frac{2\pi}{3}\)

AI Suggestion: Option D

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Pedagogical Audit
Bloom's Analysis: This is an REMEMBER question because it requires recalling the range of the principal value of the inverse cotangent function and applying a known trigonometric value.
Knowledge Dimension: FACTUAL
Justification: The question directly tests the knowledge of the principal value range of the inverse cotangent function and the values of cotangent at standard angles.
Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. It directly assesses the student's understanding of inverse trigonometric functions and their principal values, a core concept in the syllabus.
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