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Ex 3.3, 3 - Chapter 3 Class 11 Trigonometric Functions
Last updated at Feb. 12, 2020 by Teachoo
Transcript
Ex 3.3, 3 Prove that cot2 π/6 + cosec 5π/6 + 3 tan2 π/6 = 6 Taking L.H.S. cot2 π/6 + cosec 5π/6 + 3 tan2 π/6 Putting π = 180° = cot2(180/6) + cosec((5 ×180)/6) + 3 tan2(180/6) = cot2 30° + cosec (150°) + 3tan2 30° Here, tan 30° = 1/√3 cot 30° = 1/tan〖30°〗 = 1/(1/√3) = √3 For cosec 150° , First, Finding sin 150° sin 150° = sin (180 – 30° ) = sin 30° = 1/2 cosec 150° = 1/sin〖150°〗 = 1/(1/2) = 2 Putting values = cot2 30° + cosec (150°) + 3tan2 30° = (√3)2 + 2 + 3 × (1/√3)^2 = 3 + 2 + 3 × 1/3 = 3 + 2 + 1 = 6 = R.H.S Hence proved
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Chapter 3 Class 11 Trigonometric Functions
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Davneet Singh
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.