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Question Number 106387 by bemath last updated on 05/Aug/20

cot θ + cot ((π/4)+θ) = 2 θ = ?

$$\mathrm{cot}\:\theta\:+\:\mathrm{cot}\:\left(\frac{\pi}{\mathrm{4}}+\theta\right)\:=\:\mathrm{2}\: \\ $$$$\theta\:=\:? \\ $$

Answered by 1549442205PVT last updated on 05/Aug/20

cot θ + cot ((π/4)+θ) = 2 ⇔((sin(2θ+(π/4)))/(sinθsin((π/4)+θ)))=2⇔sin((π/4)+2θ)=2sinθsin((π/4)+θ) ⇔cos2θ+sin2θ=2sinθ(cosθ+sinθ) cos2θ+sin2θ=sin2θ+2sin^2 θ ⇔cos2θ=2sin^2 θ⇔cos2θ=1−cos2θ ⇔cos2θ=(1/2)=cos(π/3)⇔2θ=±(π/3)+2kπ ⇔𝛉=±(𝛑/6)+k𝛑(k∈Z)

$$\mathrm{cot}\:\theta\:+\:\mathrm{cot}\:\left(\frac{\pi}{\mathrm{4}}+\theta\right)\:=\:\mathrm{2}\: \\ $$$$\Leftrightarrow\frac{\mathrm{sin}\left(\mathrm{2}\theta+\frac{\pi}{\mathrm{4}}\right)}{\mathrm{sin}\theta\mathrm{sin}\left(\frac{\pi}{\mathrm{4}}+\theta\right)}=\mathrm{2}\Leftrightarrow\mathrm{sin}\left(\frac{\pi}{\mathrm{4}}+\mathrm{2}\theta\right)=\mathrm{2sin}\theta\mathrm{sin}\left(\frac{\pi}{\mathrm{4}}+\theta\right) \\ $$$$\Leftrightarrow\mathrm{cos2}\theta+\mathrm{sin2}\theta=\mathrm{2sin}\theta\left(\mathrm{cos}\theta+\mathrm{sin}\theta\right) \\ $$$$\mathrm{cos2}\theta+\mathrm{sin2}\theta=\mathrm{sin2}\theta+\mathrm{2sin}^{\mathrm{2}} \theta \\ $$$$\Leftrightarrow\mathrm{cos2}\theta=\mathrm{2sin}^{\mathrm{2}} \theta\Leftrightarrow\mathrm{cos2}\theta=\mathrm{1}−\mathrm{cos2}\theta \\ $$$$\Leftrightarrow\mathrm{cos2}\theta=\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{cos}\frac{\pi}{\mathrm{3}}\Leftrightarrow\mathrm{2}\theta=\pm\frac{\pi}{\mathrm{3}}+\mathrm{2k}\pi \\ $$$$\Leftrightarrow\boldsymbol{\theta}=\pm\frac{\boldsymbol{\pi}}{\mathrm{6}}+\boldsymbol{\mathrm{k}\pi}\left(\boldsymbol{\mathrm{k}}\in\mathbb{Z}\right) \\ $$

Answered by bobhans last updated on 05/Aug/20

⇒ cot ((π/4)+θ) = ((1−tan θ)/(1+tan θ )) = ((1−(1/(cot θ )))/(1+(1/(cot θ)))) = ((cot θ−1)/(cot θ+1)) . [ let cot θ = χ ] ⇒χ + ((χ−1)/(χ+1)) = 2 ; χ^2 +2χ−1=2χ+2 χ^2 = 3 ⇒ χ = ± (√3) ; cot θ = ± (√3) or tan θ = ± (1/(√3)) , θ = ± (π/6) + k.π ★

$$\Rightarrow\:\mathrm{cot}\:\left(\frac{\pi}{\mathrm{4}}+\theta\right)\:=\:\frac{\mathrm{1}−\mathrm{tan}\:\theta}{\mathrm{1}+\mathrm{tan}\:\theta\:}\:=\:\frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{cot}\:\theta\:}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{cot}\:\theta}} \\ $$$$=\:\frac{\mathrm{cot}\:\theta−\mathrm{1}}{\mathrm{cot}\:\theta+\mathrm{1}}\:.\:\left[\:\mathrm{let}\:\mathrm{cot}\:\theta\:=\:\chi\:\right]\: \\ $$$$\Rightarrow\chi\:+\:\frac{\chi−\mathrm{1}}{\chi+\mathrm{1}}\:=\:\mathrm{2}\:;\:\chi^{\mathrm{2}} +\mathrm{2}\chi−\mathrm{1}=\mathrm{2}\chi+\mathrm{2} \\ $$$$\chi^{\mathrm{2}} \:=\:\mathrm{3}\:\Rightarrow\:\chi\:=\:\pm\:\sqrt{\mathrm{3}}\:;\:\mathrm{cot}\:\theta\:=\:\pm\:\sqrt{\mathrm{3}}\: \\ $$$$\mathrm{or}\:\mathrm{tan}\:\theta\:=\:\pm\:\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\:,\:\theta\:=\:\pm\:\frac{\pi}{\mathrm{6}}\:+\:\mathrm{k}.\pi\:\bigstar \\ $$

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