Cześć!
Obliczenia
[tex]\dfrac{sin^264^o+sin^226^o-4}{sin^226^o+cos^226^o+4}=\dfrac{sin^2(90^o-26^o)+sin^226^o-4}{1+4}=\\\\\\=\dfrac{cos^226^o+sin^226^o-4}{5}=\dfrac{1-4}{5}=\boxed{-\frac{3}{5}}[/tex]
Wykorzystane wzory
[tex]sin^2\alpha+cos^2\alpha=1\\\\sin(90^o-\alpha)=cos\alpha[/tex]
