Ta có: $1-(\dfrac{sin^2x}{1+cotx}+\dfrac{cos^2x}{1-tanx})$
$=1-\dfrac{sin^2x-\dfrac{sin^3x}{cosx}+cos^2x+\dfrac{cos^3x}{sinx}}{1-1-tanx+cotx}$
$=1-\dfrac{1+\dfrac{cox^2-sin^2}{cosx.sinx}}{\dfrac{cox^2x-sin^2x}{sinx.cox}}$
$=\dfrac{sinx.cosx}{cos^2x-sin^2x}$ ???