Given #alpha=pi/3# and #beta=(2pi)/3#
We have #(beta+alpha)=piand (beta-alpha)=pi/3#
To evaluate
#(sin^2(alpha+beta)-cos^2alpha-cos^2beta)/(sin(alpha+beta)-sin^2alpha-sin^2beta)#
#=(sin^2(alpha+beta)-(cos^2alpha+cos^2beta))/(sin(alpha+beta)-(sin^2alpha+sin^2beta))#
#=(sin^2(alpha+beta)-(1-sin^2alpha+cos^2beta))/(sin(alpha+beta)-(sin^2alpha+1-cos^2beta))#
#=(sin^2(alpha+beta)-(1+cos^2beta-sin^2alpha))/(sin(alpha+beta)-(1-(cos^2beta-sin^2alpha)))#
#=(sin^2(alpha+beta)-(1+cos(beta+alpha)cos(beta-alpha)))/(sin(alpha+beta)-(1-cos(beta+alpha)cos(beta-alpha))#
#=(sin^2(pi)-(1+cos(pi)cos(pi/3)))/(sin(pi)-(1-cos(pi)cos(pi/3))#
#=(0-(1-1/2))/(0-(1+1/2))=1/3#
Formula used
#cos(beta+alpha)cos(beta-alpha)#
#=cos^2betacos^2alpha-sin^2betasin^2alpha#
#=cos^2beta(1-sin^2alpha)-(1-cos^2beta)sin^2alpha#
#=cos^2beta-sin^2alpha#
Alternative
#(sin^2(alpha+beta)-cos^2alpha-cos^2beta)/(sin(alpha+beta)-sin^2alpha-sin^2beta)#
#=(sin^2(pi)-cos^2(pi/3)-cos^2((2pi)/3))/(sin(pi)-sin^2(pi/3)-sin^2((2pi)/3))#
#=(0-1/4-1/4)/(0-3/4-3/4)=(1/2)/(3/2)=1/3#
