Last updated at May 29, 2018 by Teachoo

Transcript

Ex 2.1, 14 Find the value of tan-1 √3 – sec-1(–2) is equal to π (B) – π/3 (C) π/3 (D) 2π/3 Solving tan-1 √𝟑 Let y = tan-1 √3 tan y = √3 tan y = tan (π/3) Range of principal value of tan-1 is ((−π)/2.π/2) Hence, principal value of tan-1 √3 is π/3 Solving sec−1 (–2) Let y = sec−1 (–2) sec y = -2 sec y = sec (2π/3) We know that principal value range of sec-1 is [0,π] – {π/2} Hence, principal value of sec-1 (–2) = 2π/3 Now we have tan-1 (√3) = π/3 sec-1 (–2) = 2π/3 Solving tan-1 √3 – sec-1 (−2) = π/3 − (2π/3) = π/3 − 2π/3 = (π − 2π)/3 = - π/3 Hence , correct answer is (B)

Chapter 2 Class 12 Inverse Trigonometric Functions

Ex 2.1, 5
Important

Ex 2.1, 8 Important

Ex 2.1, 12 Important

Ex 2.1, 14 Important You are here

Example 5 Important

Example 8 Important

Ex 2.2, 12 Important

Ex 2.2, 15 Important

Ex 2.2, 19 Important

Ex 2.2, 21 Important

Example 10 Important

Example 12 Important

Example 13 Important

Misc. 2 Important

Misc. 7 Important

Misc. 10 Important

Misc. 11 Important

Misc 12 Important

Misc. 17 Important

Class 12

Important Question for exams Class 12

- Chapter 1 Class 12 Relation and Functions
- Chapter 2 Class 12 Inverse Trigonometric Functions
- Chapter 3 Class 12 Matrices
- Chapter 4 Class 12 Determinants
- Chapter 5 Class 12 Continuity and Differentiability
- Chapter 6 Class 12 Application of Derivatives
- Chapter 7 Class 12 Integrals
- Chapter 8 Class 12 Application of Integrals
- Chapter 9 Class 12 Differential Equations
- Chapter 10 Class 12 Vector Algebra
- Chapter 11 Class 12 Three Dimensional Geometry
- Chapter 12 Class 12 Linear Programming
- Chapter 13 Class 12 Probability

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.