Last updated at Feb. 12, 2020 by Teachoo
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
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