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

Transcript

Ex 3.3, 9 Prove cos (3Ο€/2+π‘₯) cos (2Ο€ + π‘₯)[cot (3Ο€/2βˆ’π‘₯) + cot (2Ο€ + π‘₯)] =1 Taking L.H.S. First we solve cos (πŸ‘π…/𝟐 "+ " 𝒙) Putting Ο€ = 180Β° = cos ((3 Γ— 180Β°)/2 " + x" ) = cos ( 270Β° + x) = cos ( 360Β° – 90Β° + x) = cos (2Ο€ – 90Β° + x) = cos (2Ο€ + (x – 90Β°)) = cos (x – 90Β°) = cos (– (90°– x)) = cos (90°– x) = sin x Now cos (2Ο€ + x) = cos x & cot (2Ο€ + x) = cot x Now we solve cot (πŸ‘π…/𝟐 "βˆ’ " 𝒙) Putting Ο€ = 180Β° = cot ((3 Γ— 180Β°)/2 " βˆ’ x" ) = cot (270Β° "βˆ’" x) = cot (360Β° – 90Β° "βˆ’" x) = cot (2Ο€ "βˆ’" 90Β° "βˆ’" x) = cot (2Ο€ "βˆ’" (x + 90Β°)) = "βˆ’"cot (x + 90Β°) = –(–tan x) = tan x Now putting values in equation cos (3Ο€/2+π‘₯) cos (2Ο€ + π‘₯)[cot (3Ο€/2βˆ’π‘₯) + cot (2Ο€ + π‘₯)] = (sin x) Γ— (cos x) Γ— [tan x + cot x] = (sin x cos x) Γ— [cot x + tan x] = (sin x cos x) Γ— [cos⁑π‘₯/sin⁑π‘₯ + sin⁑π‘₯/cos⁑π‘₯ ] = (sin x cos x) Γ— [(γ€–(cos〗⁑π‘₯) Γ— γ€–(cos〗⁑π‘₯)+γ€– (sin〗⁑π‘₯) Γ— γ€–(sin〗⁑π‘₯))/(sin⁑π‘₯ Γ— γ€–(cos〗⁑π‘₯))] = (sin x cos x) Γ— [(cos2⁑π‘₯ +γ€– sin2〗⁑π‘₯)/(sin⁑π‘₯ Γ— γ€–(cos〗⁑π‘₯))] = cos2⁑π‘₯ +γ€– sin2〗⁑π‘₯ = 1 = R.H.S Hence proved

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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.