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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)

$${Show}\:{that} \\ $$$$\frac{\mathrm{1}+{x}}{\mathrm{1}+\sqrt{\mathrm{1}+{x}}\:}\:+\frac{\mathrm{1}−{x}}{\mathrm{1}−\sqrt{\mathrm{1}−{x}\:}}\:=\mathrm{1}\:{when}\:{x}=\frac{\sqrt{\mathrm{3}\:}}{\mathrm{2}} \\ $$

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

$$\frac{\mathrm{1}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}{\mathrm{1}+\sqrt{\mathrm{1}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}}\:\:+\:\:\:\frac{\mathrm{1}−\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}{\mathrm{1}−\sqrt{\mathrm{1}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}} \\ $$$$\frac{\frac{\mathrm{2}+\sqrt{\mathrm{3}}}{\mathrm{2}}}{\mathrm{1}+\:\sqrt{\mathrm{1}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}}\:\:\:+\:\:\:\frac{\frac{\mathrm{2}−\sqrt{\mathrm{3}}}{\mathrm{2}}}{\mathrm{1}−\sqrt{\mathrm{1}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}} \\ $$$$\frac{\mathrm{2}+\sqrt{\mathrm{3}}}{\mathrm{2}\left(\mathrm{1}+\sqrt{\mathrm{1}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\right)}\:+\:\:\frac{\mathrm{2}−\sqrt{\mathrm{3}}}{\mathrm{2}\left(\mathrm{1}−\sqrt{\left.\mathrm{1}−\sqrt{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\right)}\right)} \\ $$$$\:\frac{\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\:\left(\mathrm{1}−\sqrt{\mathrm{1}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\right)}{−\sqrt{\mathrm{3}}}\:+\:\mathrm{2}+\mathrm{2}\sqrt{\mathrm{1}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:}\:−\:\sqrt{\mathrm{3}\:}\:−\sqrt{\mathrm{3}\left(\mathrm{1}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\right)}\: \\ $$$$\frac{−\mathrm{2}\sqrt{\mathrm{3}}\:+\:\mathrm{2}\sqrt{\mathrm{3}+\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}}\:−\mathrm{3}\:+\:\sqrt{\mathrm{9}+\frac{\mathrm{9}\sqrt{\mathrm{3}}}{\mathrm{2}}}\:+\:\mathrm{2}\sqrt{\mathrm{3}}\:+\:\mathrm{2}\sqrt{\mathrm{3}−\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}}\:−\mathrm{3}−\sqrt{\mathrm{9}}\:−\:\frac{\mathrm{9}\sqrt{\mathrm{3}}}{\mathrm{2}}}{\mathrm{3}} \\ $$$$\frac{\mathrm{2}\sqrt{\mathrm{3}+\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}\:}\:−\:\mathrm{6}\:+\:\sqrt{\mathrm{9}+\frac{\mathrm{9}\sqrt{\mathrm{3}}}{\mathrm{2}}\:}\:+\:\mathrm{2}\sqrt{\mathrm{3}−\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}}\:−\:\sqrt{\mathrm{9}}\:−\:\frac{\mathrm{9}\sqrt{\mathrm{3}}}{\mathrm{2}}}{\mathrm{3}} \\ $$$$\approx\:\mathrm{1} \\ $$

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

$${first}\:{find}\:\sqrt{\mathrm{1}+{x}}\:{and}\sqrt{\mathrm{1}−{x}}\: \\ $$$$\mathrm{1}+{x}=\mathrm{1}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\:\:{and}\:\mathrm{1}−{x}=\mathrm{1}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{2}+\sqrt{\mathrm{3}}}{\mathrm{2}}\:\:\:{and}\:\mathrm{1}−{x}=\frac{\mathrm{2}−\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{4}+\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{4}}\:\:\:{and}=\frac{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{4}} \\ $$$$=\frac{\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}^{\mathrm{2}} \:}\:{and}=\frac{\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}^{\mathrm{2}} } \\ $$$${so}\:\sqrt{\mathrm{1}+{x}}\:\:=\frac{\sqrt{\mathrm{3}}\:+\mathrm{1}}{\mathrm{2}\:\:}\:{and}\sqrt{\mathrm{1}−{x}}\:=\frac{\sqrt{\mathrm{3}}\:−\mathrm{1}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{2}+\sqrt{\mathrm{3}}}{\mathrm{2}\left(\mathrm{1}+\frac{\sqrt{\mathrm{3}}+\mathrm{1}}{\mathrm{2}}\right)}+\frac{\mathrm{2}−\sqrt{\mathrm{3}}}{\mathrm{2}\left(\mathrm{1}−\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}}\right)} \\ $$$$=\frac{\mathrm{2}+\sqrt{\mathrm{3}}}{\mathrm{3}+\sqrt{\mathrm{3}}}+\frac{\mathrm{2}−\sqrt{\mathrm{3}}}{\mathrm{3}−\sqrt{\mathrm{3}}} \\ $$$$=\frac{\mathrm{6}−\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{3}\sqrt{\mathrm{3}}−\mathrm{3}+\mathrm{6}+\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{3}\sqrt{\mathrm{3}}\:−\mathrm{3}}{\mathrm{9}−\mathrm{3}} \\ $$$$=\mathrm{6}/\mathrm{6}=\mathrm{1}\:{Ans} \\ $$$$ \\ $$

Commented by Rasheed.Sindhi last updated on 06/May/18

Fine!

$$\mathcal{F}{ine}! \\ $$

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