[tex]\frac{sin^2(90^o-\alpha)-cos\alpha}{(1-cos\alpha)^2-sin^2\alpha}=\frac{cos^2\alpha-cos\alpha}{1-2cos\alpha+cos^2\alpha-sin^2\alpha}=\\\\\frac{cos\alpha(cos\alpha-1)}{sin^2\alpha+cos^2\alpha-2cos\alpha+cos^2\alpha-sin^2\alpha}=\frac{cos\alpha(cos\alpha-1)}{2cos^2\alpha-2cos\alpha}=\frac{cos\alpha(cos\alpha-1)}{2cos\alpha(cos\alpha-1)}=\frac{1}{2}[/tex]
