Example 27 - Find value of tan pi/8 - Chapter 3 Class 11 - 2x 3x formula - Finding value

  1. Chapter 3 Class 11 Trigonometric Functions
  2. Serial order wise
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Example, 27 find the value of tan "Ο€" /8 . tan πœ‹/8 Putting Ο€ = 180Β° = tan (180Β°)/8 = tan (45Β°)/2 We find tan (45Β°)/2 using tan 2x formula tan 2x = (2 tan⁑π‘₯)/(1 βˆ’π‘‘π‘Žπ‘›2π‘₯) Putting x = (45Β°)/2 tan (2 Γ— (45Β°)/2) = (2 tan⁑〖 (45Β°)/2γ€—)/(1 βˆ’π‘‘π‘Žπ‘›2 (45Β°)/2) tan 45Β° = (2 tan⁑〖 (45Β°)/2γ€—)/(1 βˆ’π‘‘π‘Žπ‘›2 (45Β°)/2) 1 = (2 tan⁑〖 (45Β°)/2γ€—)/(1 βˆ’π‘‘π‘Žπ‘›2 (45Β°)/2) 1 – tan2 (45Β°)/2 = 2tan (45Β°)/2 Putting tan (45Β°)/2 = x 1 – x2 = 2x 0 = 2x + x2 – 1 0 = x2 + 2x – 1 x2 + 2x – 1 = 0 The above equation is of the form ax2 + bx + c = 0 Here a = 1, b = 2, c = -1 Solution are x = (βˆ’ 𝑏 Β± √(𝑏2 βˆ’4π‘Žπ‘) )/2π‘Ž = (βˆ’ 2 Β± √((βˆ’2)2 βˆ’ 4 Γ— 1 Γ— (βˆ’1)) )/(2 Γ— 1) = (βˆ’2 Β± √(4 + 4))/2 = (βˆ’2 Β± √8)/2 = (βˆ’2 Β± √(2 Γ— 2 Γ— 2))/2 = (βˆ’2 Β± 2√2)/2 = (2 ( βˆ’1 ±√2 ))/2 = – 1 Β± √2 β‡’ x = – 1 Β± √2 β‡’ tan (45Β°)/2 = – 1 Β± √2 But tan (45Β°)/2 = –1 – √2 is not possible as (45Β°)/2 = 22.5Β° lies in first quadrant & tan is positive in first quadrant β‡’ tan (45Β°)/2 = - 1 + √2 So, tan πœ‹/8 = √2 – 1

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