${\sin }^4\alpha +{\cos }^2\alpha +{\cos }^2\alpha {\sin }^2\alpha =$

$=\left({\sin }^4\alpha +{\cos }^2\alpha {\sin }^2\alpha \right)+{\cos }^2\alpha =$

${\sin }^2\alpha \left({\sin }^2\alpha +{\cos }^2\alpha \right)+{\cos }^2\alpha =$

$={\sin }^2\alpha +{\cos }^2\alpha =1$

Skorzystajmy z wzoru skróconego mnożenia: $a^3-b^3=(a-b)(a^2+ab+b^2)$

i zapiszmy

${\cos }^3\alpha -{\sin }^3\alpha =(\cos \alpha -\sin \alpha )({\cos }^2\alpha +\sin \alpha \cos \alpha +{\sin }^2\alpha )=$

$=(\cos \alpha -\sin \alpha )(1+\sin \alpha \cos \alpha )$.

Zatem ułamek ma postać:

$\frac{(\cos \alpha -\sin \alpha )(1+\sin \alpha \cos \alpha )}{1+\cos \alpha ·\sin \alpha }=\cos \alpha -\sin \alpha =-\frac{1}{2}$.