Question

please solve this question ..........please......​

Answer 1

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 = 1 [hence, proved]

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

1.

$\displaystyle \: tan \: (A + B) = \frac{tan \: A + tan \: B}{1 - tan \: A \: tan B }$

2.

$\displaystyle \: tan \: \frac{2\pi}{3} = tan(\pi - \frac{\pi}{3} ) = - tan \: \frac{\pi}{3} = - \sqrt{3}$

CALCULATION

$\displaystyle \: tan \: x + tan(x + \: \frac{\pi}{3}) + tan(x + \: \frac{2\pi}{3}) = 3$

$\implies \displaystyle \: \: tanx \: + \frac{tan \: x + tan \: \frac{\pi}{3} }{1 - tan \: x \: tan \frac{\pi}{3} } + \frac{tan \: x + tan \: \frac{2\pi}{3} }{1 - tan \: x \: tan \frac{2\pi}{3} } = 3$

$\implies \displaystyle \: \: tanx \: + \frac{tan \: x + \sqrt{3} }{1 - \sqrt{3} tan \: x } + \frac{tan \: x - \sqrt{3} }{1 + \sqrt{3} tan \: x \:} = 3 \: \: (by \: \: formula \: 2)$

$\implies \displaystyle \: \frac{tanx(1 - 3 {tan}^{2}x) + (tanx + \sqrt{3} ) (1 + \sqrt{3}tanx ) + \: (tanx - \sqrt{3} )(1 - \sqrt{3}tanx ) \: }{ (1 - 3 {tan}^{2}x)} = 3$

$\implies \displaystyle \: \frac{tanx - 3 {tan}^{3} x + tanx + \sqrt{3 } {tan}^{2} x + \sqrt{3} + 3tanx - \sqrt{3 } {tan}^{2} x - \sqrt{3} +</u></em><em><u>tanx</u></em><em><u>+</u></em><em><u> 3tanx }{1 - 3 {tan}^{2}x } = 3$

$\implies \: \displaystyle \: \frac{3(3tanx - {tan}^{3} x)}{1 - 3 {tan}^{2}x } = 3$

$\implies \: \displaystyle \: \frac{(3tanx - {tan}^{3} x)}{1 - 3 {tan}^{2}x } = 1$