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Example 27 - Find value of tan pi/8 - Chapter 3 Class 11

Example 27 - Chapter 3 Class 11 Trigonometric Functions - Part 2
Example 27 - Chapter 3 Class 11 Trigonometric Functions - Part 3 Example 27 - Chapter 3 Class 11 Trigonometric Functions - Part 4 Example 27 - Chapter 3 Class 11 Trigonometric Functions - Part 5

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Transcript

Example 27 find the value of tan πœ‹/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) 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 (As tan 45Β° = 1) Let tan (πŸ’πŸ“Β°)/𝟐 = x So, our equation becomes 1 – x2 = 2x 0 = 2x + x2 – 1 x2 + 2x – 1 = 0 The above equation is of the form ax2 + bx + c = 0 where a = 1, b = 2, c = βˆ’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 Thus, x = –1 Β± √2 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 Therefore, tan (45Β°)/2 = βˆ’1 + √2 i.e. tan 𝝅/πŸ– = √𝟐 – 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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.