Question

If tanx+tan(π/3+x)+tan(2π/3+x)=3 and x∈(0,π/4), then x= _____. correct answer will be marked as brainliest.

Answer 1

Step-by-step explanation:

tanx +tan(x+ pi/3) +tan(x+2pi/3) = 3.

tanx + (tanx+tanpi/3)/(1-tanx.tanpi/3) + (tanx + tan2.pi/3)/(1-tanx.tan2pi/3)=3

=> tanx + (tanx +3^1/2)/(1–3^1/2.tanx) + (tanx - 3^1/2)/(1 +3^1/2.tanx)=3

=>tanx + [ (tanx+ 3^1/2.) (1+ 3^1/2.tanx)+(tanx - 3^1/2.) (1 - 3^1/2.tanx)]/(1–3^1/2.tanx) (1+ 3^1/2.tanx)=3

=> tanx+[tanx+3^1/2+3^1/2.tan^2x +3.tanx+tanx-3^1/2–3^1/2.tan^2x+3tanx]/(1–3tan^2x).

=> tanx + 8tanx/(1–3tan^2x)=3

=> (tanx - 3tan^3x+ 8 tanx) /(1–3tan^2x)= 3

=> (9tanx -3 tan^3x) /(1–3.tan^2x) = 3

=> 3 . (3tanx - tan^3x )/(1- 3 tan^2x ) = 3

=> 3. tan3x = 3

=> tan3x = 1

=> tan 3x = tan pi /4

3x = n.pi + pi/4

x = 1/3.(n.pi + pi/4 ) , Answer

Answer 2

Answer:

Step-by-step explanation:

YOUR QUESTIONS IS--------------If tan x + tan (x + π/3) + tan (x + 2π/3) = 3

prove that tan 3x = 1

we know:- tan(A + B) = [{tanA + tanB}/{1 - tanA.tanB}]

therefore,

tanx + {tanx + tanπ/3}/{1 - tanx.tanπ/3} + {tanx + tan2π/3}/{1 - tanx.tan2π/3} = 3

=> tanx + {tanx + √3}/{1 - √3tanx} + {tanx - √3}/{1 + √3tanx} = 3

=>  = 3

=> tanx + {tanx + √3tan²x + √3 + 3tanx + tanx - √3tan²x - √3 + 3tanx}/{1 - 3tan²x} = 3

=> tanx + {8tanx}/{1 - 3tan²x} = 3

=> {tanx - 3tan³x + 8tanx}/{1 - 3tan²x} = 3

=> {9tanx - 3tan³x}/{1 - 3tan²x} = 3

=> 3{tanx - 3tan³x}/{1 - 3tan²x} = 3

we know:- tan3A = {tanA - 3tan³A}/{1 - 3tan²A}

therefore, tan3x = 3/3

=> tan3x = 3/3