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

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

$$\mathrm{if},\:\mathrm{cot}\:\theta+\mathrm{cosec}\:\theta=\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\:\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value} \\ $$ $$\mathrm{of}\:\theta\:\mathrm{where}\:\mathrm{0}<\theta\leqslant\mathrm{2}\pi. \\ $$

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)]

$$\frac{\mathrm{cos}\theta}{\mathrm{sin}\theta}+\frac{\mathrm{1}}{\mathrm{sin}\theta}=\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\Rightarrow\frac{\mathrm{1}+\mathrm{cos}\theta}{\mathrm{sin}\theta}=\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\Rightarrow \\ $$ $$\frac{\mathrm{2cos}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}}{\mathrm{2sin}\frac{\theta}{\mathrm{2}}\mathrm{cos}\frac{\theta}{\mathrm{2}}}=\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\Rightarrow\begin{cases}{\mathrm{1}.\mathrm{cos}\frac{\theta}{\mathrm{2}}=\mathrm{0}}\\{\mathrm{2}.\mathrm{cot}\frac{\theta}{\mathrm{2}}=\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}}\end{cases} \\ $$ $$\left.\mathrm{1}\right)\:\mathrm{cos}\frac{\theta}{\mathrm{2}}=\mathrm{0}\Rightarrow\frac{\theta}{\mathrm{2}}=\pm\frac{\pi}{\mathrm{2}}+\mathrm{k}\pi\Rightarrow\theta=\mathrm{k}\pi\pm\pi \\ $$ $$\left[\mathrm{for}:\mathrm{k}=\mathrm{0},\mathrm{1}\Rightarrow\theta=\mathrm{0}^{\bullet} \left[\mathrm{not}\:\mathrm{ok}\right],\pi,\mathrm{2}\pi\right] \\ $$ $$\left.\mathrm{2}\right)\mathrm{cot}\frac{\theta}{\mathrm{2}}=\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}=\mathrm{cot}\frac{\pi}{\mathrm{3}}\Rightarrow\frac{\theta}{\mathrm{2}}=\frac{\pi}{\mathrm{3}}+\mathrm{l}\pi \\ $$ $$\Rightarrow\left[\theta=\frac{\mathrm{2}\pi}{\mathrm{3}}+\mathrm{2l}\pi,\mathrm{for}:\mathrm{l}=\mathrm{0},\mathrm{1}\Rightarrow\theta=\frac{\mathrm{2}\pi}{\mathrm{3}},\frac{\mathrm{5}\pi}{\mathrm{3}}\right] \\ $$

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

cosec^2 θ−cot^2 θ=1 (cosecθ+cotθ)(cosecθ−cotθ)=1 (1/(√3))×(cosecθ−cotθ)=1 cozecθ+cotθ=(1/(√3)) cosecθ−cotθ=(√3) 2cosecθ=(1/(√3))+(√3) =(4/(√3)) cosecθ=(2/(√3))→sinθ=((√3)/2)=sin(π/3) or sin(π−(π/3))=sin((2π)/3) 2cotθ=(1/(√3))−(√3) →2cotθ=((−2)/(√3)) cotθ=((−1)/(√3))→tanθ=−(√3) =tan(π−(π/3))=tan((2π)/3) so the solution is ((2π)/3)

$${cosec}^{\mathrm{2}} \theta−{cot}^{\mathrm{2}} \theta=\mathrm{1} \\ $$ $$\left({cosec}\theta+{cot}\theta\right)\left({cosec}\theta−{cot}\theta\right)=\mathrm{1} \\ $$ $$\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}×\left({cosec}\theta−{cot}\theta\right)=\mathrm{1} \\ $$ $${cozec}\theta+{cot}\theta=\frac{\mathrm{1}}{\sqrt{\mathrm{3}}} \\ $$ $${cosec}\theta−{cot}\theta=\sqrt{\mathrm{3}}\: \\ $$ $$\mathrm{2}{cosec}\theta=\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}+\sqrt{\mathrm{3}}\:=\frac{\mathrm{4}}{\sqrt{\mathrm{3}}} \\ $$ $${cosec}\theta=\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\rightarrow{sin}\theta=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}={sin}\frac{\pi}{\mathrm{3}}\:\:{or}\:\:{sin}\left(\pi−\frac{\pi}{\mathrm{3}}\right)={sin}\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$ $$\mathrm{2}{cot}\theta=\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}−\sqrt{\mathrm{3}}\:\rightarrow\mathrm{2}{cot}\theta=\frac{−\mathrm{2}}{\sqrt{\mathrm{3}}} \\ $$ $${cot}\theta=\frac{−\mathrm{1}}{\sqrt{\mathrm{3}}}\rightarrow{tan}\theta=−\sqrt{\mathrm{3}}\:={tan}\left(\pi−\frac{\pi}{\mathrm{3}}\right)={tan}\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$ $$\boldsymbol{{so}}\:\boldsymbol{{the}}\:\boldsymbol{{solution}}\:\boldsymbol{{is}}\:\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$ $$ \\ $$

Answered by MJS last updated on 29/Oct/19

cot θ +cosec θ =(1/(√3)) (1/(tan θ))+(1/(sin θ))=(1/(√3)) ((1+cos θ)/(sin θ))=(1/(√3)) (1/(tan (θ/2)))=(1/(√3)) tan (θ/2) =(√3) θ=((2π)/3)+2nπ 0<θ≤2π ⇒ θ=((2π)/3)

$$\mathrm{cot}\:\theta\:+\mathrm{cosec}\:\theta\:=\frac{\mathrm{1}}{\sqrt{\mathrm{3}}} \\ $$ $$\frac{\mathrm{1}}{\mathrm{tan}\:\theta}+\frac{\mathrm{1}}{\mathrm{sin}\:\theta}=\frac{\mathrm{1}}{\sqrt{\mathrm{3}}} \\ $$ $$\frac{\mathrm{1}+\mathrm{cos}\:\theta}{\mathrm{sin}\:\theta}=\frac{\mathrm{1}}{\sqrt{\mathrm{3}}} \\ $$ $$\frac{\mathrm{1}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}=\frac{\mathrm{1}}{\sqrt{\mathrm{3}}} \\ $$ $$\mathrm{tan}\:\frac{\theta}{\mathrm{2}}\:=\sqrt{\mathrm{3}} \\ $$ $$\theta=\frac{\mathrm{2}\pi}{\mathrm{3}}+\mathrm{2}{n}\pi \\ $$ $$\mathrm{0}<\theta\leqslant\mathrm{2}\pi\:\Rightarrow\:\theta=\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$

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