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



Last updated at Feb. 12, 2020 by Teachoo
Learn All Concepts of Chapter 2 Class 11 Relations and Function - FREE. Check - Trigonometry Class 11 - All Concepts
Transcript
Ex 3.3, 9 Prove cos (3Ο/2+π₯) cos (2Ο + π₯)[cot (3Ο/2βπ₯) + cot (2Ο + π₯)] =1 Taking L.H.S. First we solve cos (ππ /π "+ " π) Putting Ο = 180Β° = cos ((3 Γ 180Β°)/2 " + x" ) = cos ( 270Β° + x) = cos ( 360Β° β 90Β° + x) = cos (2Ο β 90Β° + x) = cos (2Ο + (x β 90Β°)) = cos (x β 90Β°) = cos (β (90Β°β x)) = cos (90Β°β x) = sin x Now cos (2Ο + x) = cos x & cot (2Ο + x) = cot x Now we solve cot (ππ /π "β " π) Putting Ο = 180Β° = cot ((3 Γ 180Β°)/2 " β x" ) = cot (270Β° "β" x) = cot (360Β° β 90Β° "β" x) = cot (2Ο "β" 90Β° "β" x) = cot (2Ο "β" (x + 90Β°)) = "β"cot (x + 90Β°) = β(βtan x) = tan x Now putting values in equation cos (3Ο/2+π₯) cos (2Ο + π₯)[cot (3Ο/2βπ₯) + 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 [cosβ‘π₯/sinβ‘π₯ + sinβ‘π₯/cosβ‘π₯ ] = (sin x cos x) Γ [(γ(cosγβ‘π₯) Γ γ(cosγβ‘π₯)+γ (sinγβ‘π₯) Γ γ(sinγβ‘π₯))/(sinβ‘π₯ Γ γ(cosγβ‘π₯))] = (sin x cos x) Γ [(cos2β‘π₯ +γ sin2γβ‘π₯)/(sinβ‘π₯ Γ γ(cosγβ‘π₯))] = cos2β‘π₯ +γ sin2γβ‘π₯ = 1 = R.H.S Hence proved
Ex 3.3
Ex 3.3, 2 Important
Ex 3.3, 3 Important
Ex 3.3, 4
Ex 3.3, 5 Important
Ex 3.3, 6 Important
Ex 3.3, 7
Ex 3.3, 8 Important
Ex 3.3, 9 Important You are here
Ex 3.3, 10
Ex 3.3, 11 Important
Ex 3.3, 12
Ex 3.3, 13
Ex 3.3, 14
Ex 3.3, 15
Ex 3.3, 16
Ex 3.3, 17
Ex 3.3, 18 Important
Ex 3.3, 19
Ex 3.3, 20
Ex 3.3, 21 Important
Ex 3.3, 22 Important
Ex 3.3, 23 Important
Ex 3.3, 24
Ex 3.3, 25
About the Author