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

Last updated at May 29, 2023 by

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

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

 

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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

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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.