Examples

Chapter 3 Class 11 Trigonometric Functions
Serial order wise

### Transcript

Example 20 find the value of tan π/8. tan π/π Putting Ο = 180Β° = tan (180Β°)/8 = tan (ππΒ°)/π We find tan (45Β°)/2 using tan 2x formula tan 2x = (2 tanβ‘π₯)/(1 βπ‘ππ2π₯) Putting x = (ππΒ°)/π tan ("2 Γ " (45Β°)/2) = (2 tanβ‘γ (45Β°)/2γ)/(1 βπ‘ππ2 (45Β°)/2) tan 45Β° = (π πππβ‘γ (ππΒ°)/πγ)/(π βππππ (ππΒ°)/π) tan 45Β° = (2 tanβ‘γ (45Β°)/2γ)/(1 βπ‘ππ2 (45Β°)/2) 1 = (2 tanβ‘γ (45Β°)/2γ)/(1 βπ‘ππ2 (45Β°)/2) 1 β tan2 (45Β°)/2 = 2tan (45Β°)/2 Let tan (ππΒ°)/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 = β1 Solution are x = (β π Β± β(π2 β4ππ) )/2π = (β 2 Β± β((β2)2 β 4 Γ 1 Γ (β1)) )/(2 Γ 1) = (β2 Β± β(4 + 4))/2 = (βπ Β± βπ)/π = (β2 Β± β(2 Γ 2 Γ 2))/2 = (β2 Β± 2β2)/2 = (2 ( β1 Β±β2 ))/2 = β1 Β± βπ Thus, x = β1 Β± β2 tan (ππΒ°)/π = β1 Β± βπ But tan (ππΒ°)/π = β1 β βπ is not possible as (45Β°)/2 = 22.5Β° lies in first quadrant & tan is positive in first quadrant Therefore, tan (45Β°)/2 = β1 + β2 i.e. tan π/π = βπ β 1

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#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.