Let, #cos^2phi=a," so that, "sin^2phi=1-a#.
Given that, #cos^4phi/y+sin^4phi/x=1/(x+y)#,
#:. a^2/y+(1-a)^2/x=1/(x+y)#.
#:. {a^2x+(1-a)^2y}/(xy)=1/(x+y)#.
#:. {a^2x+(1-2a+a^2)y}/(xy)=1/(x+y)#.
#:. {a^2(x+y)-2ay+y}/(xy)=1/(x+y), i.e., #
# a^2(x+y)^2-2ay(x+y)+y(x+y)=xy, or,#
# a^2(x+y)^2-2ay(x+y)+y^2=0#.
#:. {a(x+y)-y}^2=0#.
# rArr a=y/(x+y), i.e., cos^2phi=a=y/(x+y)#.
#:. sin^2phi=1-a=1-y/(x+y)=x/(x+y)#.
In other words, #cos^2phi/y=1/(x+y), &, sin^2phi/x=1/(x+y)#.
Accordingly, #(cos^2phi/y)^6=1/(x+y)^6, i.e., #
#cos^12phi/y^6=1/(x+y)^6 :. cos^12phi/y^5=y/(x+y)^6#.
Similarly, #sin^12phi/x^5=x/(x+y)^6#.
Consequently, #cos^12phi/y^5+sin^12phi/x^5#,
#=y/(x+y)^6+x/(x+y)^6=1/(x+y)^6(y+x)#.
# rArr cos^12phi/y^5+sin^12phi/x^5=1/(x+y)^5#, as desired!
Feel the Joy of Maths.!
