Given that, #u_n=sin^ntheta+cos^ntheta#.
#:. u_3-u_5=(sin^3theta+cos^3theta)-(sin^5theta+cos^5theta)#,
#=(sin^3theta-sin^5theta)+(cos^3theta-cos^5theta)#,
#=sin^3theta(1-sin^2theta)+cos^3theta(1-cos^2theta)#,
#=sin^3thetacos^2theta+cos^3thetasin^2theta#,
#=sin^2thetacos^2theta(sintheta+costheta)#,
#=u_1(sin^2thetacos^2theta)#.
# rArr (u_3-u_5)/u_1=sin^2thetacos^2theta..........(ast^1)#.
Again, #u_5-u_7=sin^5theta+cos^5theta-sin^7theta-cos^7theta#,
#=sin^5theta(1-sin^2theta)+cos^5theta(1-cos^2theta)#,
#=sin^5thetacos^2theta+cos^5thetasin^2theta#,
#=sin^2thetacos^2theta(sin^3theta+cos^3theta)#,
#=u_3(sin^2thetacos^2theta)#.
# rArr (u_5-u_7)/u_3=sin^2thetacos^2theta.........(ast^2)#.
#"Hence, "(u_3-u_5)/u_1=K((u_5-u_7)/u_3), (ast^1) and (ast^2),#
# rArr K=1#.
#color(red)("Enjoy Maths.!")#
