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Ex 3.3, 9 - Prove cos (3pi/2+x) cos (2pi + x)[cot (3pi/2 - x)

Ex 3.3, 9 - Chapter 3 Class 11 Trigonometric Functions - Part 2
Ex 3.3, 9 - Chapter 3 Class 11 Trigonometric Functions - Part 3 Ex 3.3, 9 - Chapter 3 Class 11 Trigonometric Functions - Part 4

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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.