if-cot-cosec-1-3-then-find-the-value-of-where-0-lt-2pi-

Question Number 72497 by Shamim last updated on 29/Oct/19

if, \cot θ + cosec θ = \frac{1}{\sqrt{3}} then find the value

of θ where 0 < θ ⩽ 2 π .

Answered by behi83417@gmail.com last updated on 29/Oct/19

\frac{\cos θ}{\sin θ} + \frac{1}{\sin θ} = \frac{1}{\sqrt{3}} \Rightarrow \frac{1 + \cos θ}{\sin θ} = \frac{1}{\sqrt{3}} \Rightarrow

\frac{{2 \cos}^{2} \frac{θ}{2}}{2 \sin \frac{θ}{2} \cos \frac{θ}{2}} = \frac{1}{\sqrt{3}} \Rightarrow {\begin{cases} 1 . \cos \frac{θ}{2} = 0 \\ 2 . \cot \frac{θ}{2} = \frac{1}{\sqrt{3}} \end{cases}

1) \cos \frac{θ}{2} = 0 \Rightarrow \frac{θ}{2} = \pm \frac{π}{2} + k π \Rightarrow θ = k π \pm π

[for : k = 0, 1 \Rightarrow θ = 0^{∙} [not ok], π, 2 π]

2) \cot \frac{θ}{2} = \frac{1}{\sqrt{3}} = \cot \frac{π}{3} \Rightarrow \frac{θ}{2} = \frac{π}{3} + l π

\Rightarrow [θ = \frac{2 π}{3} + 2 l π, for : l = 0, 1 \Rightarrow θ = \frac{2 π}{3}, \frac{5 π}{3}]

Answered by Tanmay chaudhury last updated on 29/Oct/19

{c o s e c}^{2} θ - {c o t}^{2} θ = 1

(c o s e c θ + c o t θ) (c o s e c θ - c o t θ) = 1

\frac{1}{\sqrt{3}} \times (c o s e c θ - c o t θ) = 1

c o z e c θ + c o t θ = \frac{1}{\sqrt{3}}

c o s e c θ - c o t θ = \sqrt{3}

2 c o s e c θ = \frac{1}{\sqrt{3}} + \sqrt{3} = \frac{4}{\sqrt{3}}

c o s e c θ = \frac{2}{\sqrt{3}} \to s i n θ = \frac{\sqrt{3}}{2} = s i n \frac{π}{3} o r s i n (π - \frac{π}{3}) = s i n \frac{2 π}{3}

2 c o t θ = \frac{1}{\sqrt{3}} - \sqrt{3} \to 2 c o t θ = \frac{- 2}{\sqrt{3}}

c o t θ = \frac{- 1}{\sqrt{3}} \to t a n θ = - \sqrt{3} = t a n (π - \frac{π}{3}) = t a n \frac{2 π}{3}

s o t h e s o l u t i o n i s \frac{2 π}{3}

Answered by MJS last updated on 29/Oct/19

\cot θ + cosec θ = \frac{1}{\sqrt{3}}

\frac{1}{\tan θ} + \frac{1}{\sin θ} = \frac{1}{\sqrt{3}}

\frac{1 + \cos θ}{\sin θ} = \frac{1}{\sqrt{3}}

\frac{1}{\tan \frac{θ}{2}} = \frac{1}{\sqrt{3}}

\tan \frac{θ}{2} = \sqrt{3}

θ = \frac{2 π}{3} + 2 n π

0 < θ ⩽ 2 π \Rightarrow θ = \frac{2 π}{3}

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