Learn All Concepts of Chapter 2 Class 11 Relations and Function - FREE. Check - Trigonometry Class 11 - All Concepts

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

Transcript

Ex 3.3, 5 Find the value of: sin 75° sin 75° = sin (45° + 30°) = sin 45° cos 30° + cos 45° sin 30° = 1/√2 × √3/2 + 1/√2 × 1/2 = 1/√2 (√3/2 " + " 1/2) = (√𝟑 + 𝟏)/(𝟐√𝟐) sin (x + y) = sin x cos y + cos x sin y Putting x = 45° and y = 30° 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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.