Last updated at May 29, 2018 by Teachoo

Transcript

Ex3.4, 8 Find the general solution of the equation sec2 2x = 1 – tan 2x sec2 2x = 1 – tan 2x ⇒ 1 + tan2 2x = 1 – tan2x ⇒ tan2 2x + tan2x = 1 – 1 ⇒ tan2 2x + tan2x = 0 tan 2x (tan2x + 1)= 0 So, tan 2x = 0 or tan 2x + 1 = 0 tan 2x = 0 or tan 2x = – 1 We find solutions for tan 2x = 0 & tan 2x = – 1 separately General solution for tan 2x = 0 Let tan x = tan y ⇒ tan 2x = tan 2y Also, tan 2x = 0 From (1) and (2) tan 2y = 0 tan 2y = tan 0 ⇒ 2y = 0 ⇒ y = 0 General solution is given by 2x = nπ + 2y 2x = nπ + 0 2x = nπ x = 𝑛𝜋/2 General solution for tan 2x = -1 Let tan x = tan y ⇒ tan 2x = tan 2y Given tan 2x = – 1 From (3) and (4) tan 2y = – 1 tan 2y = tan 3/4 π ⇒ 2y = 3/4 π General solution is given by 2x = nπ + 2y where n ∈ Z Putting 2y = 3/4 π 2x = nπ + 3/4 π x = 1/2 × (nπ + 3/4 π ) x = 1/2 × (nπ + 3/4 π ) x = 𝑛𝜋/2 + 3/8 π where n ∈ Z Hence, General Solution are For tan 2x = 0 , x = 𝑛𝜋/2 For tan 2x = -1 , x = 𝑛𝜋/2 + 3/8 π Where n ∈ Z

Ex 3.1, 1
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Ex 3.1, 2 Important

Ex 3.2, 7 Important

Ex 3.2, 8 Important

Ex 3.2, 9 Important

Ex 3.3, 4 Important

Ex 3.3, 5 Important

Ex 3.3, 8 Important

Ex 3.3, 11 Important

Ex 3.3, 18 Important

Ex 3.3, 23 Important

Ex 3.3, 21 Important

Example 24 Important

Ex 3.4, 4 Important

Ex 3.4, 8 Important You are here

Ex 3.4, 9 Important

Example 27 Important

Example 28 Important

Misc 4 Important

Misc 7 Important

Class 11

Important Question for exams Class 11

- Chapter 1 Class 11 Sets
- Chapter 2 Class 11 Relations and Functions
- Chapter 3 Class 11 Trigonometric Functions
- Chapter 4 Class 11 Mathematical Induction
- Chapter 5 Class 11 Complex Numbers
- Chapter 6 Class 11 Linear Inequalities
- Chapter 7 Class 11 Permutations and Combinations
- Chapter 8 Class 11 Binomial Theorem
- Chapter 9 Class 11 Sequences and Series
- Chapter 10 Class 11 Straight Lines
- Chapter 11 Class 11 Conic Sections
- Chapter 12 Class 11 Introduction to Three Dimensional Geometry
- Chapter 13 Class 11 Limits and Derivatives
- Chapter 14 Class 11 Mathematical Reasoning
- Chapter 15 Class 11 Statistics
- Chapter 16 Class 11 Probability

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.