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

\cot θ + \cot (\frac{π}{4} + θ) = 2

θ = ?

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

\cot θ + \cot (\frac{π}{4} + θ) = 2

\Leftrightarrow \frac{\sin (2 θ + \frac{π}{4})}{\sin θ \sin (\frac{π}{4} + θ)} = 2 \Leftrightarrow \sin (\frac{π}{4} + 2 θ) = 2 \sin θ \sin (\frac{π}{4} + θ)

\Leftrightarrow \cos 2 θ + \sin 2 θ = 2 \sin θ (\cos θ + \sin θ)

\cos 2 θ + \sin 2 θ = \sin 2 θ + {2 \sin}^{2} θ

\Leftrightarrow \cos 2 θ = {2 \sin}^{2} θ \Leftrightarrow \cos 2 θ = 1 - \cos 2 θ

\Leftrightarrow \cos 2 θ = \frac{1}{2} = \cos \frac{π}{3} \Leftrightarrow 2 θ = \pm \frac{π}{3} + 2 k π

\Leftrightarrow θ = \pm \frac{π}{6} + k π (k \in Z)

Answered by bobhans last updated on 05/Aug/20

\Rightarrow \cot (\frac{π}{4} + θ) = \frac{1 - \tan θ}{1 + \tan θ} = \frac{1 - \frac{1}{\cot θ}}{1 + \frac{1}{\cot θ}}

= \frac{\cot θ - 1}{\cot θ + 1} . [let \cot θ = χ]

\Rightarrow χ + \frac{χ - 1}{χ + 1} = 2; χ^{2} + 2 χ - 1 = 2 χ + 2

χ^{2} = 3 \Rightarrow χ = \pm \sqrt{3}; \cot θ = \pm \sqrt{3}

or \tan θ = \pm \frac{1}{\sqrt{3}}, θ = \pm \frac{π}{6} + k . π ★

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