Question-190527

Question Number 190527 by Rupesh123 last updated on 04/Apr/23

Commented by Frix last updated on 05/Apr/23

$$\pm\frac{\mathrm{5}}{\mathrm{6}} \\ $$

Answered by cortano12 last updated on 05/Apr/23

⇒cos θ−sin θ=((√(13))/(12)) sin 2θ let sin 2θ = y ⇒1−y=((13)/(144)) y^2 ⇒13y^2 +144y−144=0 ⇒y=((12)/(13)) ⇒sin 2θ=((12)/(13)) ⇒tan 2θ=((2tan θ)/(1−tan^2 θ))=((12)/5) ⇒5+5tan θ=12−12tan θ ⇒12−12tan^2 θ=10tan θ ⇒6tan^2 θ+5tan θ−6=0 ⇒ { ((tan θ=(2/3))),((tan θ=−(3/2))) :}

$$\:\Rightarrow\mathrm{cos}\:\theta−\mathrm{sin}\:\theta=\frac{\sqrt{\mathrm{13}}}{\mathrm{12}}\:\mathrm{sin}\:\mathrm{2}\theta \\ $$$$\:\mathrm{let}\:\mathrm{sin}\:\mathrm{2}\theta\:=\:\mathrm{y} \\ $$$$\Rightarrow\mathrm{1}−\mathrm{y}=\frac{\mathrm{13}}{\mathrm{144}}\:\mathrm{y}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{13y}^{\mathrm{2}} +\mathrm{144y}−\mathrm{144}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{y}=\frac{\mathrm{12}}{\mathrm{13}}\:\Rightarrow\mathrm{sin}\:\mathrm{2}\theta=\frac{\mathrm{12}}{\mathrm{13}} \\ $$$$\Rightarrow\mathrm{tan}\:\mathrm{2}\theta=\frac{\mathrm{2tan}\:\theta}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\theta}=\frac{\mathrm{12}}{\mathrm{5}} \\ $$$$\Rightarrow\mathrm{5}+\mathrm{5tan}\:\theta=\mathrm{12}−\mathrm{12tan}\:\theta \\ $$$$\Rightarrow\mathrm{12}−\mathrm{12tan}\:^{\mathrm{2}} \theta=\mathrm{10tan}\:\theta \\ $$$$\Rightarrow\mathrm{6tan}\:^{\mathrm{2}} \theta+\mathrm{5tan}\:\theta−\mathrm{6}=\mathrm{0} \\ $$$$\Rightarrow\begin{cases}{\mathrm{tan}\:\theta=\frac{\mathrm{2}}{\mathrm{3}}}\\{\mathrm{tan}\:\theta=−\frac{\mathrm{3}}{\mathrm{2}}}\end{cases} \\ $$

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