Solve2+cot⁡θ=csc⁡θwhere0≤θ&lt;2π ?

We know csc2θ−cot2θ=1⇒(csc⁡θ+cot⁡θ)(csc⁡θ−cot⁡θ)=1Given csc⁡θ−cot⁡θ=2So, csc⁡θ+cot⁡θ=12So, 2csc⁡θ=52⇒sin⁡θ=45&gt;0cot⁡θ=12−csc⁡θ=12−54&lt;0So, θ will lie in the 2nd quadrant, the principal value in (π2,π)

Check the solution to "Solve2+cot⁡θ=csc⁡θwhere0≤θ<2π ?"

manhangasto6

Answered question

2022-02-26

Solve

2 + \cot θ = \csc θ

where

0 \leq θ < 2 π

Answer & Explanation

Nathan Kent

Beginner2022-02-27Added 8 answers

We know

\csc^{2} θ - \cot^{2} θ = 1

\Rightarrow (\csc θ + \cot θ) (\csc θ - \cot θ) = 1

Given

\csc θ - \cot θ = 2

So,

\csc θ + \cot θ = \frac{1}{2}

So,

2 \csc θ = \frac{5}{2} \Rightarrow \sin θ = \frac{4}{5} > 0

\cot θ = \frac{1}{2} - \csc θ = \frac{1}{2} - \frac{5}{4} < 0

So,

θ

will lie in the 2nd quadrant, the principal value in

(\frac{π}{2}, π)

Elodie Williamson

Beginner2022-02-28Added 7 answers

2 + \cot θ = \csc θ \Rightarrow 2 + \frac{\cos θ}{\sin θ} = \frac{1}{\sin θ}

\Rightarrow 2 \sin θ + \cos θ = 1 (\sin θ \neq 1)

\Rightarrow \frac{2}{\sqrt{5}} \sin θ + \frac{1}{\sqrt{5}} \cos θ = \frac{1}{\sqrt{5}}

\Rightarrow \cos a \sin θ + \sin a \cos θ = \frac{1}{\sqrt{5}}

where

a = \arctan (\frac{1}{2})

\Rightarrow \sin (a + θ) = \frac{1}{\sqrt{5}}

I think you can solve it from here.
Alternatively,
Let

\cot θ = x

, then,

\csc^{2} θ = 1 + \cot^{2} θ = 1 + x^{2}

Thus,

{(2 + \cot θ)}^{2} = \csc^{2} θ

\Rightarrow 4 + x^{2} + 4 x = x^{2} + 1

\Rightarrow 4 x = - 3 \Rightarrow x = - \frac{3}{4} \Rightarrow \tan θ = - \frac{4}{3}

\Rightarrow θ = π - \arctan (\frac{4}{3}), 2 π - \arctan (\frac{4}{3})

But only

π - \arctan (\frac{4}{3})

satisfies the original equation.Thus, the solution is

θ = π - \arctan (\frac{4}{3})

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