Last updated at March 9, 2017 by Teachoo

Transcript

Example, 27 find the value of tan "Ο" /8 . tan π/8 Putting Ο = 180Β° = tan (180Β°)/8 = tan (45Β°)/2 We find tan (45Β°)/2 using tan 2x formula tan 2x = (2 tanβ‘π₯)/(1 βπ‘ππ2π₯) Putting x = (45Β°)/2 tan (2 Γ (45Β°)/2) = (2 tanβ‘γ (45Β°)/2γ)/(1 βπ‘ππ2 (45Β°)/2) tan 45Β° = (2 tanβ‘γ (45Β°)/2γ)/(1 βπ‘ππ2 (45Β°)/2) 1 = (2 tanβ‘γ (45Β°)/2γ)/(1 βπ‘ππ2 (45Β°)/2) 1 β tan2 (45Β°)/2 = 2tan (45Β°)/2 Putting tan (45Β°)/2 = x 1 β x2 = 2x 0 = 2x + x2 β 1 0 = x2 + 2x β 1 x2 + 2x β 1 = 0 The above equation is of the form ax2 + bx + c = 0 Here a = 1, b = 2, c = -1 Solution are x = (β π Β± β(π2 β4ππ) )/2π = (β 2 Β± β((β2)2 β 4 Γ 1 Γ (β1)) )/(2 Γ 1) = (β2 Β± β(4 + 4))/2 = (β2 Β± β8)/2 = (β2 Β± β(2 Γ 2 Γ 2))/2 = (β2 Β± 2β2)/2 = (2 ( β1 Β±β2 ))/2 = β 1 Β± β2 β x = β 1 Β± β2 β tan (45Β°)/2 = β 1 Β± β2 But tan (45Β°)/2 = β1 β β2 is not possible as (45Β°)/2 = 22.5Β° lies in first quadrant & tan is positive in first quadrant β tan (45Β°)/2 = - 1 + β2 So, tan π/8 = β2 β 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

Ex 3.4, 9 Important

Example 27 Important You are here

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

CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .