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Chapter 3 Class 11 Trigonometric Functions
Ex 3.1, 2 (i)
Ex 3.2, 7 Important
Ex 3.2, 8
Ex 3.2, 9 Important
Ex 3.3, 4
Ex 3.3, 5 (i) 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
Question 7 Important Deleted for CBSE Board 2024 Exams
Question 4 Important Deleted for CBSE Board 2024 Exams
Question 8 Important Deleted for CBSE Board 2024 Exams
Question 9 Important Deleted for CBSE Board 2024 Exams
Example 20 Important You are here
Example 21 Important
Misc 4 Important
Misc 7 Important
Chapter 3 Class 11 Trigonometric Functions
Last updated at June 22, 2023 by Teachoo
Example 20 find the value of tan π/8. tan π /π Putting Ο = 180Β° = tan (180Β°)/8 = tan (ππΒ°)/π We find tan (45Β°)/2 using tan 2x formula tan 2x = (2 tanβ‘π₯)/(1 βπ‘ππ2π₯) Putting x = (ππΒ°)/π tan ("2 Γ " (45Β°)/2) = (2 tanβ‘γ (45Β°)/2γ)/(1 βπ‘ππ2 (45Β°)/2) tan 45Β° = (π πππβ‘γ (ππΒ°)/πγ)/(π βππππ (ππΒ°)/π) 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 Let tan (ππΒ°)/π = x So, our equation becomes 1 β x2 = 2x 0 = 2x + x2 β 1 x2 + 2x β 1 = 0 The above equation is of the form ax2 + bx + c = 0 where a = 1, b = 2, c = β1 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 = (βπ Β± βπ)/π = (β2 Β± β(2 Γ 2 Γ 2))/2 = (β2 Β± 2β2)/2 = (2 ( β1 Β±β2 ))/2 = β1 Β± βπ Thus, x = β1 Β± β2 tan (ππΒ°)/π = β1 Β± βπ But tan (ππΒ°)/π = β1 β βπ is not possible as (45Β°)/2 = 22.5Β° lies in first quadrant & tan is positive in first quadrant Therefore, tan (45Β°)/2 = β1 + β2 i.e. tan π /π = βπ β 1