Given
#cos^2theta=(m^2-1)/3#
#=>3cos^2theta+1=m^2.....(1)#
Again given
#tan^3(theta/2)=tanalpha#
#=>tan^2(theta/2)=tan^(2/3)alpha...(2)#
We are to prove
#sin^(2/3)alpha+cos^(2/3)alpha=(2/m)^(2/3)#
Now expanding
#(sin^(2/3)alpha+cos^(2/3)alpha)^3# we get
#(sin^(2/3)alpha+cos^(2/3)alpha)^3#
#=(sin^2alpha+cos^2alpha)+3sin^(2/3)alphacos^(2/3)alpha(sin^(2/3)alpha+cos^(2/3)alpha)#
#=1+3sin^(2/3)alphacos^(2/3)alpha(sin^(2/3)alpha+cos^(2/3)alpha)#
#=1+(3sin^(2/3)alphacos^(2/3)alpha(sin^(2/3)alpha+cos^(2/3)alpha))/((sin^2alpha+cos^2alpha)#
#=1+((3sin^(2/3)alphacos^(2/3)alpha(sin^(2/3)alpha+cos^(2/3)alpha))/cos^2alpha)/((sin^2alpha+cos^2alpha)/cos^2alpha)#
#=1+(3tan^(2/3)alpha(1+tan^(2/3)alpha))/(1+tan^2alpha)#
#=1+(3tan^(2/3)alpha(1+tan^(2/3)alpha))/((1+tan^(2/3)alpha)(1-tan^(2/3)alpha+tan^(4/3)alpha)#
#=1+(3tan^(2/3)alpha)/(1-tan^(2/3)alpha+tan^(4/3)alpha)#
#=(1-tan^(2/3)alpha+tan^(4/3)alpha+3tan^(2/3)alpha)/(1-tan^(2/3)alpha+tan^(4/3)alpha)#
#=(1+tan^(2/3)alpha)^2/(1-tan^(2/3)alpha+tan^(4/3)alpha)#
#color(red)("using relation (2)"->tan^2(theta/2)=tan^(2/3)alpha)#
#=(1+tan^2(theta/2))^2/(1-tan^2(theta/2)+tan^4(theta/2))#
#=1/(cos^4(theta/2)(1-tan^2(theta/2)+tan^4(theta/2))#
#=1/(cos^4(theta/2)-sin^2(theta/2)cos^2(theta/2)+sin^4(theta/2))#
#=1/((cos^2(theta/2)-sin^2(theta/2))^2+sin^2(theta/2)cos^2(theta/2))#
#=1/((cos^2theta+1/4*4*sin^2(theta/2)cos^2(theta/2))#
#=4/((4cos^2theta+sin^2theta)#
#=4/((4cos^2theta+1-cos^2theta)#
#=4/((3cos^2theta+1))=4/m^2=(2/m)^2->color(red)("by relation(1))"#
Hence proved
#sin^(2/3)alpha+cos^(2/3)alpha=(2/m)^(2/3)#