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Question Number 202604 by MATHEMATICSAM last updated on 30/Dec/23

If x = (((√(a + 1)) + (√(a − 1)))/( (√(a + 1)) − (√(a − 1)))) and y = (((√(a + 1)) − (√(a − 1)))/( (√(a + 1)) + (√(a − 1)))) then show that ((x^2 − xy + y^2 )/(x^2 + xy + y^2 )) = ((4a^2 − 3)/(4a^2 − 1)) .

Ifx=a+1+a−1a+1−a−1andy=a+1−a−1a+1+a−1thenshowthatx2−xy+y2x2+xy+y2=4a2−34a2−1.

Answered by Rasheed.Sindhi last updated on 30/Dec/23

x = (((√(a + 1)) + (√(a − 1)) )/( (√(a + 1)) − (√(a − 1)))) =(A/B) , y=(B/A) A+B=((√(a + 1)) + (√(a − 1)) )+((√(a + 1)) − (√(a − 1)) ) =2(√(a+1)) AB=((√(a + 1)) + (√(a − 1)) )((√(a + 1)) − (√(a − 1)) ) =((√(a + 1)) )^2 −( (√(a − 1)) )^2 =(a+1)−(a−1)=2 ((x^2 − xy + y^2 )/(x^2 + xy + y^2 )) = ((4a^2 − 3)/(4a^2 − 1)) lhs:((x^2 − xy + y^2 )/(x^2 + xy + y^2 ))=(((x+y)^2 −3xy)/((x+y)^2 −xy)) x+y=(A/B)+(B/A)=((A^2 +B^2 )/(AB))=(((A+B)^2 −2AB)/(AB)) =((( 2(√(a+1)) )^2 −2(2))/2)=2(a+1)−2=2a xy=(A/B)×(B/A)=1 =(((2a)^2 −3(1))/((2a)^2 −(1)))=((4a^2 −3)/(4a^2 −1))=rhs

x=a+1+a−1a+1−a−1=AB,y=BAA+B=(a+1+a−1)+(a+1−a−1)=2a+1AB=(a+1+a−1)(a+1−a−1)=(a+1)2−(a−1)2=(a+1)−(a−1)=2x2−xy+y2x2+xy+y2=4a2−34a2−1lhs:x2−xy+y2x2+xy+y2=(x+y)2−3xy(x+y)2−xyx+y=AB+BA=A2+B2AB=(A+B)2−2ABAB=(2a+1)2−2(2)2=2(a+1)−2=2axy=AB×BA=1=(2a)2−3(1)(2a)2−(1)=4a2−34a2−1=rhs

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