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

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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) = ((√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 14 years. He provides courses for Maths, Science and Computer Science at Teachoo