Inverse Trigonometric Functions
Oswaal
CBSE
Grade 12
Question:
Principal value of $\cot^{-1}(-\sqrt{3})$ is:
(a) $5\pi/6$
(b) $-\pi/6$
(c) $\pi/6$
(d) $2\pi/3$
Step-by-Step Solution
Key Concept: cot^{-1}(-\sqrt{3}) = \pi - \pi/6 = 5\pi/6.
The detailed step-by-step mathematical proof is available inside the Mathbee app workspace.
Correct Answer: $5\pi/6$
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