1. Chapter 3 Class 11 Trigonometric Functions (Term 2)
  2. Serial order wise
  3. Ex 3.3

Ex 3.3, 5 (ii) - Chapter 3 Class 11 Trigonometric Functions (Term 2)

Last updated at Aug. 27, 2021 by

Ex 3.3, 5 - Chapter 3 Class 11 Trigonometric Functions - Part 2

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Ex 3.3, 5 - Chapter 3 Class 11 Trigonometric Functions - Part 3

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Ex 3.3, 5 - Chapter 3 Class 11 Trigonometric Functions - Part 4

 

  1. Chapter 3 Class 11 Trigonometric Functions (Term 2)
  2. Serial order wise

    3. Ex 3.3

    Ex 3.4 →

Transcript

Ex 3.3, 5 Find the value of: (ii) tan 15° tan 15° = tan (45° – 30°) = (tan 45° − 〖 tan〗⁡〖30°〗)/(1 + tan 45°tan⁡〖30°〗 ) = (1 − 1/√3)/(1 + 1 × 1/√3) Using tan (x − y) = (𝑡𝑎𝑛𝑥 − 𝑡𝑎𝑛⁡𝑦)/(1+𝑡𝑎𝑛 𝑥 𝑡𝑎𝑛⁡𝑦 ) Putting x = 45° and y = 30° = ((√3 − 1" " )/√3)/((√3 + 1" " )/√3) = (√3 −1)/√3 × √3/(√3 + 1) = (√3 − 1)/(√3 + 1) Rationalizing = (√3 − 1)/(√3 + 1) × (√3 − 1)/(√3 − 1) = (√3 − 1)2/(√3 + 1)(√3 − 1) Using (a – b)2 = a2 + b2 – 2ab = ((√3)2 + 12 − 2" " × √3 × 1)/(√3 + 1)(√3 − 1) = (3 + 1 − 2√3)/(√(3 )+ 1)(√3 − 1) Using (a – b ) (a + b) = a2 – b2 = (4 − 2√3)/((√3)2 − (1)2) = (4 − 2√3)/(3 − 1) = (2 (2 − √(3 )))/2 = 2 – √3 Hence, tan 15° = 2 – √𝟑

Chapter 3 Class 11 Trigonometric Functions (Term 2)
Serial order wise

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.