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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

BeMath°dx°(1)Given(x+1+x2)(y+1+y2)=1find(x+y)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

JSsetu=x+1+x2u.(y+1+y2)=1y+1+y2=1u1+y2=1uy1+y2=(1uy)21+y2=1u22yu+y21=1u22yu;2yu=1u212y=1uuy=12(1uu)1u=1x+1+x2×x1+x2x1+x21u=x1+x2x2(1+x2)=1+x2xy=12((1+x2x)(x+1+x2))y=12(2x)=xtherefore(x+y)2=02=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?

Fromgivenconditionwehave(x+1+x2)2(y+1+y2)2=1(x+y)2=(x1+x2+y1+y2)(x+y)2=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+1+x2=1+y2yx+y=1+y21+x2x2+y2+2xy=1+y2+1+x221+x2+y2+x2y21xy=1+x2+y2+x2y22xy=x2+y2(x+y)2=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

(x+1+x2)(y+1+y2)=1(1)Putx=tanA,y=tanB(A,B(π2;π2))Since1+tan2θ=1cos2θ,wehave(1)(tanA+1cosA)(tanB+1cosB)=1(1+sinA)(1+sinB)=cosAcosB1+sinA+sinB+sinAsinB=cosAcosB(2)PuttanA2=m,tanB2=n,sincesinθ=2tanθ21+tan2θandcosθ=1tan2θ21+tan2θ2,so(2)1+2m1+m2+2n1+n2+4mn(1+m2)(1+n2)=(1m2)(1n2)(1+m2)(1+n2)(1+m2)(1+n2)+2m(1+n2)+2n(1+m2)+4mn=1m2n2+m2n22m2+2n2+2(m+n)+2mn(m+n)+4mn=02(m+n)2+2(m+n)(1+mn)=02(m+n)(1+m+n+mn)=02(m+n)(1+m)(1+n)=0If1+m=0tanA2=1A2=π4A=π2thisiscontradicriontothehypothesis,so1+m0Similarly,1+n0.Hencewemusthavem+n=0tanA2+tanB2=0tan(A+B2)1tanA2tanB2=0tanA+B2=0A+B=kπx+y=tanA+tanB=tan(A+B)1tanAtanB=01tanAtanB=0(x+y)2=0.Thus,(x+y)2=0secondway(2)cosAcosBsinAsinB=1+sinA+sinBcos(A+B)=1+2sinA+B2cosAB212sin2A+B2=1+2sinA+B2cosAB22sinA+B2(cosAB2+1)=0(3)SinceA,B(π2;π2),AB2πcosAB21cosAB2+10Hence(3)sinA+B2=0A+B2=0A+B=0tan(A+B)=0tanA+tanB=tan(A+B)1tanAtanB=0Therefore,(x+y)2=(tanA+tanB)2=0

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