{ ((tan (α+2β)=(√(1+2k)))),((tan^2 (α+β){1+k tan^2 β}=k+tan^2 β)) :} Find cot 2β .


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Question Number 192537 by cortano12 last updated on 20/May/23
${ ((tan (α+2β)=(√(1+2k)))),((tan^2 (α+β){1+k tan^2 β}=k+tan^2 β)) :} Find cot 2β .$
${\begin{cases} \tan (α + 2 β) = \sqrt{1 + 2 k} \\ \tan^{2} (α + β) {1 + k \tan^{2} β} = k + \tan^{2} β \end{cases}$ $Find \cot 2 β .$
	Answered by a.lgnaoui last updated on 20/May/23
	$tan [(α+β)+β]=(√(1+2k)) • ((tan (α+β)+tan β)/(1−tan (α+β)tan β))=(√(1+2k)) (1) 2•tan^2 (α+β)=((k+tan^2 β)/(1+ktan^2 β)) (2) (1)⇔ tan (α+β)+tan β=(√(1+2k)) −tan βtan (α+β)(√(1+2k )) tan (α+β)+((√(1+2k)) )tan β×tan (α+β) =(√(1+2k)) −tan β tan (α+β)=(((√(1+2k)) −tan β)/(1+tan β(√(1+2k )))) (3) tan^2 (α+β)=((k+tan^2 β)/(1+ktan^2 β)) (4) posons y=tan (𝛂+β) x=tan 𝛃 { ((y=(((√(1+2k)) −x)/(1+x(√(1+2k)))) )),((y^2 =((k+x^2 )/(1+kx^2 )))) :} ⇒ ((1+2k+x^2 −2x(√(1+2k)))/(1+x^2 (1+2k)+2x(√(1+2k))))=((k+x^2 )/(1+kx^2 )) 1+2k+x^2 −2x(√(1+2k)) + kx^2 +2k^2 x^2 +kx^4 −(2k(√(1+2k)) ) x^3 = =kx^4 −2k((√(1+2k)) )x^3 +(2k^2 +k+1)x^2 −2x(√(1+2k)) +2k+1 =k+(1+2k)kx^2 +2kx(√(1+2k)) + x^2 +(1+2k)x^4 +2x^3 (√(1+2k)) =(1+2k)x^4 +2(√(1+2k)) x^3 +(1+k+2k^2 )x^2 +2k(√(1+2k)) x+k ⇒(1+k)x^4 +2(1+k)(√(1+2k)) )x^3 +2(k+1)(√(1+2k)) x−(k+1)=0 x^4 +2(√(1+2k)) x^3 +2x(√(1+2k)) −1=0 (x^2 +x(√(1+2k)) )^2 −[x^2 (1+2k)−2x(√(1+2k))) +1]=0 x^2 (x+(√(1+2k)) )^2 −[(x(√(1+2k)) )−1]^2 =0 ⇒x(x+(√(1+2k)) )=1−x(√(1+2k)) x^2 +2x(√(1+2k)) −1=0 (x+(√(1+2k)) )^2 =2(1+k) x=(√(2(1+k))) −(√(1+2k)) ⇒tan 𝛃=(√(2(1+k))) −(√(1+2k)) ⇒tan 2𝛃=((2tan 𝛃)/(1−tan^2 𝛃)) =((2((√(2(1+k))) −(√(1+2k)) ))/(1−[4k+3−2(√(2(1+k)(1+2k))))) cot 𝛃=((2(√(2(1+k)(1+2k))) −(4k+2))/(2((√(2(1+k))) −(√(1+2k)) ))) cot 𝛃=(((√(2(1+k)(1+2k)) )−(2k+1))/( (√(2(1+k))) −(√(1+2k)))) = (((1+2k)((√(2(1+k))) −(√(2k+1)) +(√(1+2k)) ))/1) conclusion: cot 𝛃=(1+2k)(√(2(1+k))) { (),() :}$
	$\tan [(α + β) + β] = \sqrt{1 + 2 k}$ $∙ \frac{\tan (α + β) + \tan β}{1 - \tan (α + β) \tan β} = \sqrt{1 + 2 k} (1)$ $2 ∙ \tan^{2} (α + β) = \frac{k + \tan^{2} β}{1 + ktan^{2} β} (2)$ $(1) \Leftrightarrow \tan (α + β) + \tan β = \sqrt{1 + 2 k} - \tan β \tan (α + β) \sqrt{1 + 2 k}$ $\tan (α + β) + (\sqrt{1 + 2 k}) \tan β \times \tan (α + β)$ $= \sqrt{1 + 2 k} - \tan β$ $\tan (α + β) = \frac{\sqrt{1 + 2 k} - \tan β}{1 + \tan β \sqrt{1 + 2 k}} (3)$ $\tan^{2} (α + β) = \frac{k + \tan^{2} β}{1 + ktan^{2} β} (4)$ $posons y = \tan (α + β) x = \tan β$ ${\begin{cases} y = \frac{\sqrt{1 + 2 k} - x}{1 + x \sqrt{1 + 2 k}} \\ y^{2} = \frac{k + x^{2}}{1 + {kx}^{2}} \end{cases}$ $\Rightarrow \frac{1 + 2 k + x^{2} - 2 x \sqrt{1 + 2 k}}{1 + x^{2} (1 + 2 k) + 2 x \sqrt{1 + 2 k}} = \frac{k + x^{2}}{1 + {kx}^{2}}$ $1 + 2 k + x^{2} - 2 x \sqrt{1 + 2 k} +$ ${kx}^{2} + 2 k^{2} x^{2} + {kx}^{4} - (2 k \sqrt{1 + 2 k}) x^{3} =$ $= {kx}^{4} - 2 k (\sqrt{1 + 2 k}) x^{3} + (2 k^{2} + k + 1) x^{2}$ $- 2 x \sqrt{1 + 2 k} + 2 k + 1$ $= k + (1 + 2 k) {kx}^{2} + 2 kx \sqrt{1 + 2 k} +$ $x^{2} + (1 + 2 k) x^{4} + 2 x^{3} \sqrt{1 + 2 k}$ $= (1 + 2 k) x^{4} + 2 \sqrt{1 + 2 k} x^{3} + (1 + k + 2 k^{2}) x^{2} + 2 k \sqrt{1 + 2 k} x + k$ $\Rightarrow (1 + k) x^{4} + 2 (1 + k) \sqrt{1 + 2 k}) x^{3} + 2 (k + 1) \sqrt{1 + 2 k} x - (k + 1) = 0$ $x^{4} + 2 \sqrt{1 + 2 k} x^{3} + 2 x \sqrt{1 + 2 k} - 1 = 0$ ${(x^{2} + x \sqrt{1 + 2 k})}^{2} - [x^{2} (1 + 2 k) - 2 x \sqrt{1 + 2 k)} + 1] = 0$ $x^{2} {(x + \sqrt{1 + 2 k})}^{2} - {[(x \sqrt{1 + 2 k}) - 1]}^{2} = 0$ $\Rightarrow x (x + \sqrt{1 + 2 k}) = 1 - x \sqrt{1 + 2 k}$ $x^{2} + 2 x \sqrt{1 + 2 k} - 1 = 0$ ${(x + \sqrt{1 + 2 k})}^{2} = 2 (1 + k)$ $x = \sqrt{2 (1 + k)} - \sqrt{1 + 2 k}$ $\Rightarrow \tan β = \sqrt{2 (1 + k)} - \sqrt{1 + 2 k}$ $\Rightarrow \tan 2 β = \frac{2 \tan β}{1 - \tan^{2} β} = \frac{2 (\sqrt{2 (1 + k)} - \sqrt{1 + 2 k})}{1 - [4 k + 3 - 2 \sqrt{2 (1 + k) (1 + 2 k)}}$ $\cot β = \frac{2 \sqrt{2 (1 + k) (1 + 2 k)} - (4 k + 2)}{2 (\sqrt{2 (1 + k)} - \sqrt{1 + 2 k})}$ $\cot β = \frac{\sqrt{2 (1 + k) (1 + 2 k}) - (2 k + 1)}{\sqrt{2 (1 + k)} - \sqrt{1 + 2 k}}$ $= \frac{(1 + 2 k) (\sqrt{2 (1 + k)} - \sqrt{2 k + 1} + \sqrt{1 + 2 k})}{1}$ $conclusion :$ $\cot β = (1 + 2 k) \sqrt{2 (1 + k)}$ ${\begin{cases} \end{cases}$
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