Solve all your doubts with Teachoo Black (new monthly pack available now!)
Ex 3.3
Ex 3.3, 2 Important
Ex 3.3, 3 Important
Ex 3.3, 4
Ex 3.3, 5 (i) Important
Ex 3.3, 5 (ii)
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 Important
Ex 3.3, 14
Ex 3.3, 15
Ex 3.3, 16 Important
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
Ex 3.3
Last updated at Feb. 12, 2020 by Teachoo
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