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Last updated at May 12, 2021 by Teachoo

Transcript

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 We know that cot−1 (−x) = 𝜋 − cot −1 x Since cot 𝜋/3 = 1/√3 𝜋/3 = cot−1 (𝟏/√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

Examples

Example 1
Important

Example 2 You are here

Example 3 (i) Important

Example 3 (ii)

Example 4 Deleted for CBSE Board 2022 Exams

Example 5 Important Deleted for CBSE Board 2022 Exams

Example 6 Important Deleted for CBSE Board 2022 Exams

Example 7 Deleted for CBSE Board 2022 Exams

Example 8 Deleted for CBSE Board 2022 Exams

Example 9 Important

Example 10 Important Deleted for CBSE Board 2022 Exams

Example 11 Important Deleted for CBSE Board 2022 Exams

Example 12 Important Deleted for CBSE Board 2022 Exams

Example 13 Important Deleted for CBSE Board 2022 Exams

Chapter 2 Class 12 Inverse Trigonometric Functions (Term 1)

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.