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Question Number 34422 by math1967 last updated on 06/May/18
Show that ((1+x)/(1+(√(1+x)) )) +((1−x)/(1−(√(1−x )))) =1 when x=((√(3 ))/2)
Showthat1+x1+1+x+1−x1−1−x=1whenx=32
Answered by Rio Mike last updated on 06/May/18
((1+((√3)/2))/(1+(√(1+((√3)/2))))) + ((1− ((√3)/2))/(1−(√(1−((√3)/2))))) (((2+(√3))/2)/(1+ (√(1+((√3)/2))))) + (((2−(√3))/2)/(1−(√(1−((√3)/2))))) ((2+(√3))/(2(1+(√(1+((√3)/2)))))) + ((2−(√3))/(2(1−(√(1−(√((√3)/2)))))))) (((2+(√3)) (1−(√(1+((√3)/2)))))/(−(√3))) + 2+2(√(1−((√3)/2) )) − (√(3 )) −(√(3(1−((√3)/2) ))) ((−2(√3) + 2(√(3+((3(√3))/2))) −3 + (√(9+((9(√3))/2))) + 2(√3) + 2(√(3−((3(√3))/2))) −3−(√9) − ((9(√3))/2))/3) ((2(√(3+((3(√3))/2) )) − 6 + (√(9+((9(√3))/2) )) + 2(√(3−((3(√3))/2))) − (√9) − ((9(√3))/2))/3) ≈ 1
1+321+1+32+1−321−1−322+321+1+32+2−321−1−322+32(1+1+32)+2−32(1−1−32))(2+3)(1−1+32)−3+2+21−32−3−3(1−32)−23+23+332−3+9+932+23+23−332−3−9−932323+332−6+9+932+23−332−9−9323≈1
Answered by tanmay.chaudhury50@gmail.com last updated on 06/May/18
first find (√(1+x)) and(√(1−x)) 1+x=1+((√3)/2) and 1−x=1−((√3)/2) =((2+(√3))/2) and 1−x=((2−(√3))/2) =((4+2(√3))/4) and=((4−2(√3))/4) =((((√3)+1)^2 )/(2^2 )) and=((((√3)−1)^2 )/2^2 ) so (√(1+x)) =(((√3) +1)/(2 )) and(√(1−x)) =(((√3) −1)/2) =((2+(√3))/(2(1+(((√3)+1)/2))))+((2−(√3))/(2(1−(((√3)−1)/2)))) =((2+(√3))/(3+(√3)))+((2−(√3))/(3−(√3))) =((6−2(√3)+3(√3)−3+6+2(√3)−3(√3) −3)/(9−3)) =6/6=1 Ans
firstfind1+xand1−x1+x=1+32and1−x=1−32=2+32and1−x=2−32=4+234and=4−234=(3+1)222and=(3−1)222so1+x=3+12and1−x=3−12=2+32(1+3+12)+2−32(1−3−12)=2+33+3+2−33−3=6−23+33−3+6+23−33−39−3=6/6=1Ans
Commented by Rasheed.Sindhi last updated on 06/May/18
Fine!
Fine!

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