if, cot θ+cosec θ=(1/(√3)) then find the value of θ where 0<θ≤2π.


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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
	$((cosθ)/(sinθ))+(1/(sinθ))=(1/(√3))⇒((1+cosθ)/(sinθ))=(1/(√3))⇒ ((2cos^2 (θ/2))/(2sin(θ/2)cos(θ/2)))=(1/(√3))⇒ { ((1.cos(θ/2)=0)),((2.cot(θ/2)=(1/(√3)))) :} 1) cos(θ/2)=0⇒(θ/2)=±(π/2)+kπ⇒θ=kπ±π [for:k=0,1⇒θ=0^• [not ok],π,2π] 2)cot(θ/2)=(1/(√3))=cot(π/3)⇒(θ/2)=(π/3)+lπ ⇒[θ=((2π)/3)+2lπ,for:l=0,1⇒θ=((2π)/3),((5π)/3)]$
	$\frac{\cos θ}{\sin θ} + \frac{1}{\sin θ} = \frac{1}{\sqrt{3}} \Rightarrow \frac{1 + \cos θ}{\sin θ} = \frac{1}{\sqrt{3}} \Rightarrow$ $\frac{{2 \cos}^{2} \frac{θ}{2}}{2 \sin \frac{θ}{2} \cos \frac{θ}{2}} = \frac{1}{\sqrt{3}} \Rightarrow {\begin{cases} 1 . \cos \frac{θ}{2} = 0 \\ 2 . \cot \frac{θ}{2} = \frac{1}{\sqrt{3}} \end{cases}$ $1) \cos \frac{θ}{2} = 0 \Rightarrow \frac{θ}{2} = \pm \frac{π}{2} + k π \Rightarrow θ = k π \pm π$ $[for : k = 0, 1 \Rightarrow θ = 0^{∙} [not ok], π, 2 π]$ $2) \cot \frac{θ}{2} = \frac{1}{\sqrt{3}} = \cot \frac{π}{3} \Rightarrow \frac{θ}{2} = \frac{π}{3} + l π$ $\Rightarrow [θ = \frac{2 π}{3} + 2 l π, for : l = 0, 1 \Rightarrow θ = \frac{2 π}{3}, \frac{5 π}{3}]$
	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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