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Question Number 144727 by imjagoll last updated on 28/Jun/21
Given { ((m=cos θ−sin θ)),((n=cos θ+sin θ)) :} then (√(m/n)) +(√(n/m)) = ?
$$\:\:\mathrm{Given}\:\begin{cases}{\mathrm{m}=\mathrm{cos}\:\theta−\mathrm{sin}\:\theta}\\{\mathrm{n}=\mathrm{cos}\:\theta+\mathrm{sin}\:\theta}\end{cases} \\ $$$$\:\:\mathrm{then}\:\sqrt{\frac{\mathrm{m}}{\mathrm{n}}}\:+\sqrt{\frac{\mathrm{n}}{\mathrm{m}}}\:=\:? \\ $$
Answered by liberty last updated on 28/Jun/21
⇔(√(m/n))+(√(n/m))= ((m+n)/( (√(mn)))) { ((m^2 =cos^2 −sin 2θ+sin^2 θ)),((n^2 =cos^2 θ+sin 2θ+sin^2 θ)) :} { ((m^2 =1−sin 2θ)),((n^2 =1+sin 2θ)) :}⇔m^2 +n^2 =2 ⇒(m+n)^2 −2mn=2 ⇒m+n=(√(2+2mn)) ⇒m.n=cos^2 θ−sin^2 θ=cos 2θ then (√(m/n))+(√(n/m))=((√(2+2cos 2θ))/( (√(cos 2θ)))) = (√(2sec 2θ+2))
$$\:\Leftrightarrow\sqrt{\frac{\mathrm{m}}{\mathrm{n}}}+\sqrt{\frac{\mathrm{n}}{\mathrm{m}}}=\:\frac{\mathrm{m}+\mathrm{n}}{\:\sqrt{\mathrm{mn}}} \\ $$$$\begin{cases}{\mathrm{m}^{\mathrm{2}} =\mathrm{cos}\:^{\mathrm{2}} −\mathrm{sin}\:\mathrm{2}\theta+\mathrm{sin}\:^{\mathrm{2}} \theta}\\{\mathrm{n}^{\mathrm{2}} =\mathrm{cos}\:^{\mathrm{2}} \theta+\mathrm{sin}\:\mathrm{2}\theta+\mathrm{sin}\:^{\mathrm{2}} \theta}\end{cases} \\ $$$$\begin{cases}{\mathrm{m}^{\mathrm{2}} =\mathrm{1}−\mathrm{sin}\:\mathrm{2}\theta}\\{\mathrm{n}^{\mathrm{2}} =\mathrm{1}+\mathrm{sin}\:\mathrm{2}\theta}\end{cases}\Leftrightarrow\mathrm{m}^{\mathrm{2}} +\mathrm{n}^{\mathrm{2}} =\mathrm{2} \\ $$$$\Rightarrow\left(\mathrm{m}+\mathrm{n}\right)^{\mathrm{2}} −\mathrm{2mn}=\mathrm{2} \\ $$$$\Rightarrow\mathrm{m}+\mathrm{n}=\sqrt{\mathrm{2}+\mathrm{2mn}} \\ $$$$\Rightarrow\mathrm{m}.\mathrm{n}=\mathrm{cos}\:^{\mathrm{2}} \theta−\mathrm{sin}\:^{\mathrm{2}} \theta=\mathrm{cos}\:\mathrm{2}\theta \\ $$$$\mathrm{then}\:\sqrt{\frac{\mathrm{m}}{\mathrm{n}}}+\sqrt{\frac{\mathrm{n}}{\mathrm{m}}}=\frac{\sqrt{\mathrm{2}+\mathrm{2cos}\:\mathrm{2}\theta}}{\:\sqrt{\mathrm{cos}\:\mathrm{2}\theta}} \\ $$$$\:=\:\sqrt{\mathrm{2sec}\:\mathrm{2}\theta+\mathrm{2}}\: \\ $$$$ \\ $$
Commented by Rasheed.Sindhi last updated on 28/Jun/21
Typo sir! { ((m^2 =1−sin 2θ)),((n^2 =1+sin 2θ)) :}⇔m^2 +n^2 =2
$$\mathcal{T}{ypo}\:{sir}! \\ $$$$\begin{cases}{\mathrm{m}^{\mathrm{2}} =\mathrm{1}−\mathrm{sin}\:\mathrm{2}\theta}\\{\mathrm{n}^{\mathrm{2}} =\mathrm{1}+\mathrm{sin}\:\mathrm{2}\theta}\end{cases}\Leftrightarrow\mathrm{m}^{\mathrm{2}} +\mathrm{n}^{\mathrm{2}} =\mathrm{2}\:\:\: \\ $$$$ \\ $$
Commented by liberty last updated on 28/Jun/21
yes sir. thank you
$$\mathrm{yes}\:\mathrm{sir}.\:\mathrm{thank}\:\mathrm{you} \\ $$
Answered by Olaf_Thorendsen last updated on 28/Jun/21
m = cosθ−sinθ n = cosθ+sinθ x = (√(m/n))+(√(n/m)) x^2 = (m/n)+2+(n/m) x^2 = ((m^2 +n^2 )/(mn))+2 x^2 = (((1−sin2θ)+(1+sin2θ))/(cos^2 θ−sin^2 θ))+2 x^2 = ((2+2cos2θ)/(cos2θ)) x^2 = ((4cos^2 θ)/(cos2θ)) x = ((2∣cosθ∣)/( (√(cos2θ))))
$${m}\:=\:\mathrm{cos}\theta−\mathrm{sin}\theta \\ $$$${n}\:=\:\mathrm{cos}\theta+\mathrm{sin}\theta \\ $$$${x}\:=\:\sqrt{\frac{{m}}{{n}}}+\sqrt{\frac{{n}}{{m}}} \\ $$$${x}^{\mathrm{2}} \:=\:\frac{{m}}{{n}}+\mathrm{2}+\frac{{n}}{{m}} \\ $$$${x}^{\mathrm{2}} \:=\:\frac{{m}^{\mathrm{2}} +{n}^{\mathrm{2}} }{{mn}}+\mathrm{2} \\ $$$${x}^{\mathrm{2}} \:=\:\frac{\left(\mathrm{1}−\mathrm{sin2}\theta\right)+\left(\mathrm{1}+\mathrm{sin2}\theta\right)}{\mathrm{cos}^{\mathrm{2}} \theta−\mathrm{sin}^{\mathrm{2}} \theta}+\mathrm{2} \\ $$$${x}^{\mathrm{2}} \:=\:\frac{\mathrm{2}+\mathrm{2cos2}\theta}{\mathrm{cos2}\theta} \\ $$$${x}^{\mathrm{2}} \:=\:\frac{\mathrm{4cos}^{\mathrm{2}} \theta}{\mathrm{cos2}\theta} \\ $$$${x}\:=\:\frac{\mathrm{2}\mid\mathrm{cos}\theta\mid}{\:\sqrt{\mathrm{cos2}\theta}} \\ $$

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