Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

Question Number 115345 by bemath last updated on 25/Sep/20

sec θ (sec θ (sin^2 θ)+2(√3) sin θ)=1 has the roots are θ_1 and θ_2 . Find the value of tan θ_1 ×tan θ_2 .

$$\mathrm{sec}\:\theta\:\left(\mathrm{sec}\:\theta\:\left(\mathrm{sin}\:^{\mathrm{2}} \theta\right)+\mathrm{2}\sqrt{\mathrm{3}}\:\mathrm{sin}\:\theta\right)=\mathrm{1} \\ $$$${has}\:{the}\:{roots}\:{are}\:\theta_{\mathrm{1}} \:{and}\:\theta_{\mathrm{2}} .\:{Find}\:{the} \\ $$$${value}\:{of}\:\mathrm{tan}\:\theta_{\mathrm{1}} ×\mathrm{tan}\:\theta_{\mathrm{2}} . \\ $$

Answered by bobhans last updated on 25/Sep/20

⇔ ((sin^2 θ)/(cos θ)) + 2(√3) sin θ = cos θ ⇔ 2(√3) sin θ.cos θ = cos 2θ ⇒ (√3) sin 2θ = cos 2θ → tan 2θ = (1/( (√3))) →((2tan θ)/(1−tan^2 θ)) = (1/( (√3))) ⇒ tan^2 θ+2(√3) tan θ−1=0 just apply Vieta′s rule we get tan θ_1 ×tan θ_2 = −1

$$\Leftrightarrow\:\frac{\mathrm{sin}\:^{\mathrm{2}} \theta}{\mathrm{cos}\:\theta}\:+\:\mathrm{2}\sqrt{\mathrm{3}}\:\mathrm{sin}\:\theta\:=\:\mathrm{cos}\:\theta \\ $$$$\Leftrightarrow\:\mathrm{2}\sqrt{\mathrm{3}}\:\mathrm{sin}\:\theta.\mathrm{cos}\:\theta\:=\:\mathrm{cos}\:\mathrm{2}\theta\: \\ $$$$\Rightarrow\:\sqrt{\mathrm{3}}\:\mathrm{sin}\:\mathrm{2}\theta\:=\:\mathrm{cos}\:\mathrm{2}\theta\:\rightarrow\:\mathrm{tan}\:\mathrm{2}\theta\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$$\rightarrow\frac{\mathrm{2tan}\:\theta}{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \theta}\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:\Rightarrow\:\mathrm{tan}\:^{\mathrm{2}} \theta+\mathrm{2}\sqrt{\mathrm{3}}\:\mathrm{tan}\:\theta−\mathrm{1}=\mathrm{0} \\ $$$${just}\:{apply}\:{Vieta}'{s}\:{rule}\:{we}\:{get}\: \\ $$$$\mathrm{tan}\:\theta_{\mathrm{1}} ×\mathrm{tan}\:\theta_{\mathrm{2}} \:=\:−\mathrm{1} \\ $$

Answered by Dwaipayan Shikari last updated on 25/Sep/20

((sin^2 θ)/(cos^2 θ))+((2(√3)sinθ)/(cosθ))=1 tan^2 θ+2(√3)tanθ−1=0 tanθ=((−2(√3)±(√(12+4)))/2)=2−(√3) or −(2+(√3)) tanθ_1 =2−(√3) tanθ_2 =−(2+(√3)) tanθ_1 .tanθ_2 =−1

$$\frac{\mathrm{sin}^{\mathrm{2}} \theta}{\mathrm{cos}^{\mathrm{2}} \theta}+\frac{\mathrm{2}\sqrt{\mathrm{3}}\mathrm{sin}\theta}{\mathrm{cos}\theta}=\mathrm{1} \\ $$$$\mathrm{tan}^{\mathrm{2}} \theta+\mathrm{2}\sqrt{\mathrm{3}}\mathrm{tan}\theta−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{tan}\theta=\frac{−\mathrm{2}\sqrt{\mathrm{3}}\pm\sqrt{\mathrm{12}+\mathrm{4}}}{\mathrm{2}}=\mathrm{2}−\sqrt{\mathrm{3}}\:\:\mathrm{or}\:−\left(\mathrm{2}+\sqrt{\mathrm{3}}\right) \\ $$$$\mathrm{tan}\theta_{\mathrm{1}} =\mathrm{2}−\sqrt{\mathrm{3}} \\ $$$$\mathrm{tan}\theta_{\mathrm{2}} =−\left(\mathrm{2}+\sqrt{\mathrm{3}}\right) \\ $$$$\mathrm{tan}\theta_{\mathrm{1}} .\mathrm{tan}\theta_{\mathrm{2}} =−\mathrm{1} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Answered by MJS_new last updated on 25/Sep/20

(1/c)((s^2 /c)+2(√3)s)=1 t^2 +2(√3)t−1=0 ⇒ t_1 ×t_2 =−1

$$\frac{\mathrm{1}}{{c}}\left(\frac{{s}^{\mathrm{2}} }{{c}}+\mathrm{2}\sqrt{\mathrm{3}}{s}\right)=\mathrm{1} \\ $$$${t}^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{3}}{t}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\:{t}_{\mathrm{1}} ×{t}_{\mathrm{2}} =−\mathrm{1} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com