Observe that, #sqrt(2+sqrt3)=sqrt(2+2sqrt(3/4))={2+2sqrt(3/4)}^(1/2)#,
#={(3/2+1/2)+2*sqrt(3/2)*sqrt(1/2)}^(1/2)#,
#={(sqrt(3/2))^2+(sqrt(1/2))^2+2*sqrt(3/2)*sqrt(1/2)}^(1/2)#,
#=[{(sqrt(3/2)+sqrt(1/2)}^2]^(1/2)#,
#=(sqrt(3/2)+sqrt(1/2))#.
#:. sqrt2+sqrt(2+sqrt3)=sqrt2+(sqrt(3/2)+sqrt(1/2))#,
#=(2+sqrt3+1)/sqrt2#,
#rArr sqrt2+sqrt(2+sqrt3)=(3+sqrt3)/sqrt2#.
Hence, #x=(2+sqrt3)/(sqrt2+sqrt(2+sqrt3))#,
#=(2+sqrt3)*sqrt2/(3+sqrt3)#,
#=(2+sqrt3)*sqrt2/(3+sqrt3)xx(3-sqrt3)/(3-sqrt3)#,
#={(2+sqrt3)sqrt2(3-sqrt3)}/6#
#={(2+sqrt3)sqrt2*sqrt3(sqrt3-1)}/6#,
#=sqrt6/6(3+sqrt3-2)#.
#rArr x=sqrt6/6(1+sqrt3)#.
Similarly, #y=(2-sqrt3)/(sqrt2-sqrt(2-sqrt3))#,
#rArry=sqrt6/6(sqrt3-1)#.
#:." The Reqd. Value"=x+y#,
#=sqrt6/6{(1+sqrt3)+(sqrt3-1)}#,
#=sqrt2#.
