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Question Number 108291 by bemath last updated on 16/Aug/20

((∥ BeMath ∥)/(°∫ dx°)) (1) Given (x+(√(1+x^2 )))(y+(√(1+y^2 )))=1 find (x+y)^2

$$\:\:\:\frac{\parallel\:\mathcal{B}{e}\mathcal{M}{ath}\:\parallel}{°\int\:{dx}°} \\ $$$$\left(\mathrm{1}\right)\:{Given}\:\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)\left({y}+\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }\right)=\mathrm{1} \\ $$$${find}\:\left({x}+{y}\right)^{\mathrm{2}} \: \\ $$

Answered by john santu last updated on 16/Aug/20

((⇈ JS ⇈)/♥) ⇔ set u = x+(√(1+x^2 )) ⇒u.(y+(√(1+y^2 ))) = 1 y+(√(1+y^2 )) = (1/u) ⇒(√(1+y^2 )) = (1/u)−y 1+y^2 = ((1/u)−y)^2 1+y^2 =(1/u^2 )−((2y)/u)+y^2 1=(1/u^2 )−((2y)/u) ; ((2y)/u)= (1/u^2 )−1 2y=(1/u)−u ⇒ y=(1/2)((1/u)−u) (1/u)=(1/(x+(√(1+x^2 )))) ×((x−(√(1+x^2 )))/(x−(√(1+x^2 )))) (1/u) = ((x−(√(1+x^2 )))/(x^2 −(1+x^2 )))=(√(1+x^2 ))−x y=(1/2)(((√(1+x^2 ))−x)−(x+(√(1+x^2 )))) y=(1/2)(−2x)=−x therefore (x+y)^2 =0^2 =0

$$\:\:\:\:\:\frac{\upuparrows\:{JS}\:\upuparrows}{\heartsuit} \\ $$$$\Leftrightarrow\:{set}\:{u}\:=\:{x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$\Rightarrow{u}.\left({y}+\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }\right)\:=\:\mathrm{1}\: \\ $$$${y}+\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{{u}}\:\Rightarrow\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{{u}}−{y} \\ $$$$\mathrm{1}+{y}^{\mathrm{2}} \:=\:\left(\frac{\mathrm{1}}{{u}}−{y}\right)^{\mathrm{2}} \\ $$$$\mathrm{1}+{y}^{\mathrm{2}} =\frac{\mathrm{1}}{{u}^{\mathrm{2}} }−\frac{\mathrm{2}{y}}{{u}}+{y}^{\mathrm{2}} \\ $$$$\mathrm{1}=\frac{\mathrm{1}}{{u}^{\mathrm{2}} }−\frac{\mathrm{2}{y}}{{u}}\:;\:\frac{\mathrm{2}{y}}{{u}}=\:\frac{\mathrm{1}}{{u}^{\mathrm{2}} }−\mathrm{1}\: \\ $$$$\mathrm{2}{y}=\frac{\mathrm{1}}{{u}}−{u}\:\Rightarrow\:{y}=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{{u}}−{u}\right) \\ $$$$\frac{\mathrm{1}}{{u}}=\frac{\mathrm{1}}{{x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:×\frac{{x}−\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{{x}−\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \\ $$$$\frac{\mathrm{1}}{{u}}\:=\:\frac{{x}−\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{{x}^{\mathrm{2}} −\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}=\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }−{x} \\ $$$${y}=\frac{\mathrm{1}}{\mathrm{2}}\left(\left(\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }−{x}\right)−\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)\right) \\ $$$${y}=\frac{\mathrm{1}}{\mathrm{2}}\left(−\mathrm{2}{x}\right)=−{x} \\ $$$${therefore}\:\left({x}+{y}\right)^{\mathrm{2}} =\mathrm{0}^{\mathrm{2}} =\mathrm{0} \\ $$

Commented by bemath last updated on 16/Aug/20

nice!!

$${nice}!! \\ $$

Commented by udaythool last updated on 17/Aug/20

From given condition we have (x+(√(1+x^2 )))^2 (y+(√(1+y^2 )))^2 =1 ⇒(x+y)^2 =−(x(√(1+x^2 ))+y(√(1+y^2 ))) ⇒(x+y)^2 =0?

$$\mathrm{From}\:\mathrm{given}\:\mathrm{condition}\:\mathrm{we}\:\mathrm{have} \\ $$$$\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)^{\mathrm{2}} \left({y}+\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\Rightarrow\left({x}+{y}\right)^{\mathrm{2}} =−\left({x}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }+{y}\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }\right) \\ $$$$\Rightarrow\left({x}+{y}\right)^{\mathrm{2}} =\mathrm{0}? \\ $$

Commented by udaythool last updated on 17/Aug/20

x+(√(1+x^2 )) =(√(1+y^2 ))−y ⇒x+y=(√(1+y^2 ))−(√(1+x^2 )) ⇒x^2 +y^2 +2xy=1+y^2 +1+x^2 −2(√(1+x^2 +y^2 +x^2 y^2 )) ⇒1−xy=(√(1+x^2 +y^2 +x^2 y^2 )) ⇒−2xy=x^2 +y^2 ⇒(x+y)^2 =0

$${x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} \:}\:=\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }−{y} \\ $$$$\Rightarrow{x}+{y}=\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }−\sqrt{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$\Rightarrow{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{2}{xy}=\mathrm{1}+{y}^{\mathrm{2}} +\mathrm{1}+{x}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{x}^{\mathrm{2}} {y}^{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{1}−{xy}=\sqrt{\mathrm{1}+{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{x}^{\mathrm{2}} {y}^{\mathrm{2}} } \\ $$$$\Rightarrow−\mathrm{2}{xy}={x}^{\mathrm{2}} +{y}^{\mathrm{2}} \\ $$$$\Rightarrow\left({x}+{y}\right)^{\mathrm{2}} =\mathrm{0} \\ $$

Answered by 1549442205PVT last updated on 16/Aug/20

(x+(√(1+x^2 )))(y+(√(1+y^2 )))=1(1) Put x=tanA,y=tanB (A,B∈(−(π/2);(π/2))) Since 1+tan^2 θ=(1/(cos^2 θ)) ,we have (1)⇔(tanA+(1/(cosA)))(tanB+(1/(cosB)))=1 ⇔(1+sinA)(1+sinB)=cosAcosB ⇔1+sinA+sinB+sinAsinB=cosAcosB(2) Put tan(A/2)=m,tan(B/2)=n,since sinθ=((2tan(θ/2))/(1+tan^2 θ)) and cosθ=((1−tan^2 (θ/2))/(1+tan^2 (θ/2))),so (2)⇔1+((2m)/(1+m^2 ))+((2n)/(1+n^2 ))+((4mn)/((1+m^2 )(1+n^2 )))=(((1−m^2 )(1−n^2 ))/((1+m^2 )(1+n^2 ))) ⇔(1+m^2 )(1+n^2 )+2m(1+n^2 )+2n(1+m^2 )+4mn=1−m^2 −n^2 +m^2 n^2 ⇔2m^2 +2n^2 +2(m+n)+2mn(m+n)+4mn=0 ⇔2(m+n)^2 +2(m+n)(1+mn)=0 ⇔2(m+n)(1+m+n+mn)=0 ⇔2(m+n)(1+m)(1+n)=0 If 1+m=0⇒tan(A/2)=−1⇒(A/2)=((−π)/4) ⇒A=((−π)/2) this is contradicrion to the hypothesis ,so 1+m≠0 Similarly,1+n≠0.Hence we must have m+n=0 ⇒ tan(A/2)+tan(B/2)=0 ⇒((tan(((A+B)/2)))/(1−tan(A/2)tan(B/2)))=0⇒tan((A+B)/2)=0 ⇒A+B=kπ⇒x+y=tanA+tanB =((tan(A+B))/(1−tanAtanB))=(0/(1−tanAtanB))=0 ⇒(x+y)^2 =0.Thus,(x+y)^2 =0 second way (2)⇔cosAcosB−sinAsinB=1+sinA+sinB ⇔cos(A+B)=1+2sin((A+B)/2)cos((A−B)/2) ⇔1−2sin^2 ((A+B)/2)=1+2sin((A+B)/2)cos((A−B)/2) ⇔2sin((A+B)/2)(cos((A−B)/2)+1)=0(3) Since A,B∈(−(π/2);(π/2)),((A−B)/2)≠π ⇒cos ((A−B)/2)≠−1⇒cos((A−B)/2)+1≠0 Hence (3)⇔sin((A+B)/2)=0⇒((A+B)/2)=0 ⇒A+B=0⇒tan(A+B)=0 ⇒tanA+tanB=((tan(A+B))/(1−tanAtanB))=0 Therefore, (x+y)^2 =(tanA+tanB)^2 =0

$$\:\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)\left({y}+\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }\right)=\mathrm{1}\left(\mathrm{1}\right) \\ $$$$\mathrm{Put}\:\mathrm{x}=\mathrm{tanA},\mathrm{y}=\mathrm{tanB}\:\left(\mathrm{A},\mathrm{B}\in\left(−\frac{\pi}{\mathrm{2}};\frac{\pi}{\mathrm{2}}\right)\right) \\ $$$$\mathrm{Since}\:\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \theta=\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \theta}\:,\mathrm{we}\:\mathrm{have} \\ $$$$\left(\mathrm{1}\right)\Leftrightarrow\left(\mathrm{tanA}+\frac{\mathrm{1}}{\mathrm{cosA}}\right)\left(\mathrm{tanB}+\frac{\mathrm{1}}{\mathrm{cosB}}\right)=\mathrm{1} \\ $$$$\Leftrightarrow\left(\mathrm{1}+\mathrm{sinA}\right)\left(\mathrm{1}+\mathrm{sinB}\right)=\mathrm{cosAcosB} \\ $$$$\Leftrightarrow\mathrm{1}+\mathrm{sinA}+\mathrm{sinB}+\mathrm{sinAsinB}=\mathrm{cosAcosB}\left(\mathrm{2}\right) \\ $$$$\mathrm{Put}\:\mathrm{tan}\frac{\mathrm{A}}{\mathrm{2}}=\mathrm{m},\mathrm{tan}\frac{\mathrm{B}}{\mathrm{2}}=\mathrm{n},\mathrm{since}\: \\ $$$$\mathrm{sin}\theta=\frac{\mathrm{2tan}\frac{\theta}{\mathrm{2}}}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \theta}\:\mathrm{and}\:\mathrm{cos}\theta=\frac{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}},\mathrm{so} \\ $$$$\left(\mathrm{2}\right)\Leftrightarrow\mathrm{1}+\frac{\mathrm{2m}}{\mathrm{1}+\mathrm{m}^{\mathrm{2}} }+\frac{\mathrm{2n}}{\mathrm{1}+\mathrm{n}^{\mathrm{2}} }+\frac{\mathrm{4mn}}{\left(\mathrm{1}+\mathrm{m}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{n}^{\mathrm{2}} \right)}=\frac{\left(\mathrm{1}−\mathrm{m}^{\mathrm{2}} \right)\left(\mathrm{1}−\mathrm{n}^{\mathrm{2}} \right)}{\left(\mathrm{1}+\mathrm{m}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{n}^{\mathrm{2}} \right)} \\ $$$$\Leftrightarrow\left(\mathrm{1}+\mathrm{m}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{n}^{\mathrm{2}} \right)+\mathrm{2m}\left(\mathrm{1}+\mathrm{n}^{\mathrm{2}} \right)+\mathrm{2n}\left(\mathrm{1}+\mathrm{m}^{\mathrm{2}} \right)+\mathrm{4mn}=\mathrm{1}−\mathrm{m}^{\mathrm{2}} −\mathrm{n}^{\mathrm{2}} +\mathrm{m}^{\mathrm{2}} \mathrm{n}^{\mathrm{2}} \\ $$$$\Leftrightarrow\mathrm{2m}^{\mathrm{2}} +\mathrm{2n}^{\mathrm{2}} +\mathrm{2}\left(\mathrm{m}+\mathrm{n}\right)+\mathrm{2mn}\left(\mathrm{m}+\mathrm{n}\right)+\mathrm{4mn}=\mathrm{0} \\ $$$$\Leftrightarrow\mathrm{2}\left(\mathrm{m}+\mathrm{n}\right)^{\mathrm{2}} +\mathrm{2}\left(\mathrm{m}+\mathrm{n}\right)\left(\mathrm{1}+\mathrm{mn}\right)=\mathrm{0} \\ $$$$\Leftrightarrow\mathrm{2}\left(\mathrm{m}+\mathrm{n}\right)\left(\mathrm{1}+\mathrm{m}+\mathrm{n}+\mathrm{mn}\right)=\mathrm{0} \\ $$$$\Leftrightarrow\mathrm{2}\left(\mathrm{m}+\mathrm{n}\right)\left(\mathrm{1}+\mathrm{m}\right)\left(\mathrm{1}+\mathrm{n}\right)=\mathrm{0} \\ $$$$\mathrm{If}\:\mathrm{1}+\mathrm{m}=\mathrm{0}\Rightarrow\mathrm{tan}\frac{\mathrm{A}}{\mathrm{2}}=−\mathrm{1}\Rightarrow\frac{\mathrm{A}}{\mathrm{2}}=\frac{−\pi}{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{A}=\frac{−\pi}{\mathrm{2}}\:\:\mathrm{this}\:\mathrm{is}\:\mathrm{contradicrion} \\ $$$$\mathrm{to}\:\mathrm{the}\:\mathrm{hypothesis}\:,\mathrm{so}\:\mathrm{1}+\mathrm{m}\neq\mathrm{0} \\ $$$$\mathrm{Similarly},\mathrm{1}+\mathrm{n}\neq\mathrm{0}.\mathrm{Hence}\:\mathrm{we}\:\mathrm{must}\:\mathrm{have} \\ $$$$\mathrm{m}+\mathrm{n}=\mathrm{0}\:\Rightarrow\:\mathrm{tan}\frac{\mathrm{A}}{\mathrm{2}}+\mathrm{tan}\frac{\mathrm{B}}{\mathrm{2}}=\mathrm{0} \\ $$$$\Rightarrow\frac{\mathrm{tan}\left(\frac{\mathrm{A}+\mathrm{B}}{\mathrm{2}}\right)}{\mathrm{1}−\mathrm{tan}\frac{\mathrm{A}}{\mathrm{2}}\mathrm{tan}\frac{\mathrm{B}}{\mathrm{2}}}=\mathrm{0}\Rightarrow\mathrm{tan}\frac{\mathrm{A}+\mathrm{B}}{\mathrm{2}}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{A}+\mathrm{B}=\mathrm{k}\pi\Rightarrow\mathrm{x}+\mathrm{y}=\mathrm{tanA}+\mathrm{tanB} \\ $$$$=\frac{\mathrm{tan}\left(\mathrm{A}+\mathrm{B}\right)}{\mathrm{1}−\mathrm{tanAtanB}}=\frac{\mathrm{0}}{\mathrm{1}−\mathrm{tanAtanB}}=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{2}} =\mathrm{0}.\mathrm{Thus},\left(\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\boldsymbol{\mathrm{second}}\:\boldsymbol{\mathrm{way}} \\ $$$$\left(\mathrm{2}\right)\Leftrightarrow\mathrm{cosAcosB}−\mathrm{sinAsinB}=\mathrm{1}+\mathrm{sinA}+\mathrm{sinB} \\ $$$$\Leftrightarrow\mathrm{cos}\left(\mathrm{A}+\mathrm{B}\right)=\mathrm{1}+\mathrm{2sin}\frac{\mathrm{A}+\mathrm{B}}{\mathrm{2}}\mathrm{cos}\frac{\mathrm{A}−\mathrm{B}}{\mathrm{2}} \\ $$$$\Leftrightarrow\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \frac{\mathrm{A}+\mathrm{B}}{\mathrm{2}}=\mathrm{1}+\mathrm{2sin}\frac{\mathrm{A}+\mathrm{B}}{\mathrm{2}}\mathrm{cos}\frac{\mathrm{A}−\mathrm{B}}{\mathrm{2}} \\ $$$$\Leftrightarrow\mathrm{2sin}\frac{\mathrm{A}+\mathrm{B}}{\mathrm{2}}\left(\mathrm{cos}\frac{\mathrm{A}−\mathrm{B}}{\mathrm{2}}+\mathrm{1}\right)=\mathrm{0}\left(\mathrm{3}\right) \\ $$$$\mathrm{Since}\:\mathrm{A},\mathrm{B}\in\left(−\frac{\pi}{\mathrm{2}};\frac{\pi}{\mathrm{2}}\right),\frac{\mathrm{A}−\mathrm{B}}{\mathrm{2}}\neq\pi \\ $$$$\Rightarrow\mathrm{cos}\:\frac{\mathrm{A}−\mathrm{B}}{\mathrm{2}}\neq−\mathrm{1}\Rightarrow\mathrm{cos}\frac{\mathrm{A}−\mathrm{B}}{\mathrm{2}}+\mathrm{1}\neq\mathrm{0} \\ $$$$\mathrm{Hence}\:\left(\mathrm{3}\right)\Leftrightarrow\mathrm{sin}\frac{\mathrm{A}+\mathrm{B}}{\mathrm{2}}=\mathrm{0}\Rightarrow\frac{\mathrm{A}+\mathrm{B}}{\mathrm{2}}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{A}+\mathrm{B}=\mathrm{0}\Rightarrow\mathrm{tan}\left(\mathrm{A}+\mathrm{B}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{tanA}+\mathrm{tanB}=\frac{\mathrm{tan}\left(\mathrm{A}+\mathrm{B}\right)}{\mathrm{1}−\mathrm{tanAtanB}}=\mathrm{0} \\ $$$$\mathrm{Therefore},\:\left(\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}\right)^{\mathrm{2}} =\left(\boldsymbol{\mathrm{tanA}}+\boldsymbol{\mathrm{tanB}}\right)^{\mathrm{2}} =\mathrm{0} \\ $$

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