tan (x+(π/4))+3(tan (π/9)+tan ((2π)/9))=tan (x+(π/4))tan (π/9)tan ((2π)/9)


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Question Number 147643 by bobhans last updated on 22/Jul/21

$\tan (x + \frac{π}{4}) + 3 (\tan \frac{π}{9} + \tan \frac{2 π}{9}) = \tan (x + \frac{π}{4}) \tan \frac{π}{9} \tan \frac{2 π}{9}$
	Answered by liberty last updated on 22/Jul/21

	$\tan (x + \frac{π}{4}) - \tan (x + \frac{π}{4}) \tan \frac{π}{9} \tan \frac{2 π}{9} = - 3 (\tan \frac{π}{9} + \tan \frac{2 π}{9})$ $\tan (x + \frac{π}{4}) [1 - \tan \frac{π}{9} \tan \frac{2 π}{9}] = - 3 (\tan \frac{π}{9} + \tan \frac{2 π}{9})$ $\tan (x + \frac{π}{4}) = - 3 (\frac{\tan \frac{π}{9} + \tan \frac{2 π}{9}}{1 - \tan \frac{π}{9} \tan \frac{2 π}{9}})$ $\tan (x + \frac{π}{4}) = - 3 \tan \frac{π}{3} = - 3 \sqrt{3}$ $\Rightarrow x + \frac{π}{4} = - \arctan (3 \sqrt{3}) + n π$ $\Rightarrow x = - \frac{π}{4} + n π - \arctan (3 \sqrt{3})$
	Answered by gsk2684 last updated on 22/Jul/21
	$tan (x+(π/4))[1−tan (π/9)tan ((2π)/9)]=−3(tan (π/9)+tan ((2π)/9)) tan (x+(π/4))=−3(((tan (π/9)+tan ((2π)/9))/(1−tan (π/9)tan ((2π)/9)))) tan (x+(π/4))=−3 tan ((π/9)+((2π)/9)) ((tan x + 1)/(1− tan x)) =−3 tan (π/3) ((tan x + 1)/(1− tan x)) =−3(√3) tan x +1 = −3(√3) + 3(√3) tan x 3(√3)+1 = (3(√3) −1)tan x (3(√3)+1)^2 ={(3(√3))^2 −1} tan x 28+6(√(3 )) = 26 tan x ((14+3(√3))/(13))=tan x x=tan^(−1) ((14+3(√3))/(13))$
	$\tan (x + \frac{π}{4}) [1 - \tan \frac{π}{9} \tan \frac{2 π}{9}] = - 3 (\tan \frac{π}{9} + \tan \frac{2 π}{9})$ $\tan (x + \frac{π}{4}) = - 3 (\frac{\tan \frac{π}{9} + \tan \frac{2 π}{9}}{1 - \tan \frac{π}{9} \tan \frac{2 π}{9}})$ $\tan (x + \frac{π}{4}) = - 3 \tan (\frac{π}{9} + \frac{2 π}{9})$ $\frac{\tan x + 1}{1 - \tan x} = - 3 \tan \frac{π}{3}$ $\frac{\tan x + 1}{1 - \tan x} = - 3 \sqrt{3}$ $\tan x + 1 = - 3 \sqrt{3} + 3 \sqrt{3} \tan x$ $3 \sqrt{3} + 1 = (3 \sqrt{3} - 1) \tan x$ ${(3 \sqrt{3} + 1)}^{2} = {{(3 \sqrt{3})}^{2} - 1} \tan x$ $28 + 6 \sqrt{3} = 26 \tan x$ $\frac{14 + 3 \sqrt{3}}{13} = \tan x$ $x = \tan^{- 1} \frac{14 + 3 \sqrt{3}}{13}$
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