Example 2 - Chapter 2 Class 12 Inverse Trigonometric Functions

Last updated at April 16, 2024 by

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Example 2 (Method 1) Find the principal value of cot–1 ((−1)/√3) Let y = cot−1 ((−1)/√3) y = 𝜋 − cot−1 (1/√3) y = 𝜋 − 𝛑/𝟑 y = 𝟐𝛑/𝟑 Since range of cot−1 is (0, π) Hence, Principal Value is 𝟐𝛑/𝟑 Example 2 (Method 2) Find the principal value of cot–1 ((−1)/√3) Let y = cot−1 ((−1)/√3) cot y = (−1)/√3 cot y = cot (𝟐𝛑/𝟑) Range of principal value of cot−1 is (0, π) Hence, Principal Value is 𝟐𝛑/𝟑 Rough We know that cot 60° = 𝟏/√𝟑 θ = 60° = 60 × 𝜋/180 = 𝜋/3 Since (−1)/√3 is negative Principal value is π – θ i.e. π – 𝜋/3 = 𝟐𝝅/𝟑

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.