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Applied to √3, principal cot⁻¹ returns, in radians,
Aπ/6
Bπ/3
Cπ/4
D5π/6
Answer & Solution
Correct answer: A. π/6
cot(π/6) = √3 and π/6 lies in the range (0, π) of cot⁻¹.
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