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Question

# Solve the following equations: (i) tan x+tan 2x+tan 3x=0 (ii) tan x+tan 2x=tan 3x (iii) tan 3x+tan x=2tan 2x

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Solution

## (i) We have: tan x + tan 2x + tan 3x = 0Now,tanx+ tan2x + tan (x+ 2x) = 0⇒tanx + tan2x + tan x+ tan 2x1 - tan x tan 2x = 0⇒ (tanx + tan2x) (1 - tanxtan2x) + tanx + tan2x= 0⇒(tanx + tan2x) (2 - tanx tan2x) = 0⇒ tan x+ tan 2x= 0 or 2 - tanx tan2x = 0Now, tanx + tan2x = 0 ⇒ tanx = - tan2x⇒ tanx = tan -2x⇒ x = nπ - 2x ⇒ 3x = nπ ⇒ x= nπ3, n∈ZAnd,2 - tanx tan2x = 0 ⇒ tanx tan2x = 2 ⇒ sinx cosx sin2x cos2x = 2⇒ 2 sin2x cosx cosx = 2 cos2x - 2 sin2x⇒ 4 sin2x= 2 cos2x ⇒ tan2x=12⇒ tan2x = tan2α ⇒x = mπ + α, m ∈ Z, α = tan-1 12∴ x= nπ3, n∈Z or x = mπ + α, m∈Z Here,α= tan-112(ii) Given:tanx+ tan2x = tan3xNow, tanx+ tan2x = tan (x + 2x)⇒tanx + tan 2x = tanx + tan2x1- tanx tan2x⇒ tanx+ tan2x - tanx + tan2x1- tanx tan2x = 0⇒ (tanx + tan2x) (1- tanx tan2x) - (tanx + tan2x) = 0⇒ (tanx+ tan 2x) (1- tanx tan2x -1) = 0⇒ (tanx + tan2x) (- tanx tan2x) = 0⇒ tan x + tan 2x= 0 or tanx tan2x= 0Now,tan x+ tan 2x = 0 ⇒ tan x = - tan 2x ⇒ tan x= tan -2x⇒ x= nπ - 2x, n∈Z⇒ 3x = nπ ⇒x= nπ3, n∈ZAnd,tanx tan2x = 0 ⇒ sinxcosx sin2xcos2x = 0 ⇒ 2 sin2xcos2x - sin2x = 0 ⇒ sin2x =0⇒ sin2x = sin20 ⇒ x = mπ, m∈Z∴ x= nπ3, n∈Z or x= mπ, m∈Z(iii) Given: tan3x+ tanx= 2 tan2xNow, tan3x - tan2x= tan2x- tanx⇒ tanx (1 + tan3x tan2x) = tanx(1 + tan2x tanx) (tan(A−B)=tanA−tanB1+tanAtanB) ⇒tanx (1 + tan3xtan2x - 1 - tan2x tanx)= 0⇒tanx tan2x (tan3x - tanx)= 0 ⇒ tan 2x = 0 or, tan x= 0 or, tan3x- tanx = 0And, tan 2x = 0 ⇒ 2x = nπ ⇒ x = nπ2, n∈ZOr, tan 3x - tan x = 0 ⇒ tan 3x= tan x ⇒ 3x= nπ + x⇒ 2x=nπ ⇒ x = nπ2, n∈ZAnd, tanx = 0 ⇒ x = mπ, m∈Z∴ x= nπ2, n∈Z or x = mπ, m∈Z

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