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

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Ex 3.3, 2 Prove that 2sin2 π/6 + cosec2 7π/6 cos2 π/3 = 3/2 Taking L.H.S 2sin2 π/6 + cosec2 7π/6 cos2 π/3 Putting π = 180° = 2 sin2 180/6 + cosec2 (7 ×180)/6 cos2 180/3 = 2sin2 30° + cosec2 210° cos2 60° = 2(sin 30°)2 + (cosec 210°)2 (cos 60°)2 Here, sin 30° = 1/2 & cos 60° = 1/2 For cosec 210° , lets first calculate sin 210° sin 210° = sin (180 + 30) = −sin 30° = (−1)/2 So, cosec 210° = 1/sin⁡〖210°〗 = 1/((−1)/2) = 2/(−1 ) = –2 (Using sin (180 + θ) = –sin θ ) Putting values in equation = 2(sin 30°)2 + (cosec 210°)2 (cos 60°)2 = 2 (1/2)^2 + (–2)2 × (1/2)^2 = 2 × 1/4 + 4 × 1/4 = 1/2 + 1 = 1/2 + 1 = 3/2 = R.H.S Hence proved

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.