Mamy:
[tex]\dfrac{2\cos\alpha}{1-\sin\alpha}+\dfrac{2\cos\alpha}{1+\sin\alpha}=5\\\\\\\dfrac{2\cos\alpha(1+\sin\alpha)+2\cos\alpha(1-\sin\alpha)}{(1-\sin\alpha)(1+\sin\alpha)}=5\\\\\\\dfrac{2\cos\alpha+2\cos\alpha\sin\alpha+2\cos\alpha-2\cos\alpha\sin\alpha}{1-\sin^2\alpha}=5\\\\\\\dfrac{4\cos\alpha}{\cos^2\alpha}=5\\\\\\\dfrac{4}{\cos\alpha}=5\\\\\\\cos\alpha=\dfrac{4}{5}[/tex]
Korzystamy z jedynki trygonometrycznej:
[tex]\cos^2\alpha+\sin^2\alpha=1\\\\\Big(\dfrac{4}{5}\Big)^2+\sin^2\alpha=1\\\\\dfrac{16}{25}+\sin^2\alpha=1\\\\\sin^2\alpha=\dfrac{9}{25}\\\\\sin\alpha=\dfrac{3}{5}[/tex]
Bierzemy dodatnią wartość, ponieważ kąt jest ostry. Stąd:
[tex]\sin\alpha-\cos\alpha=\dfrac{3}{5}-\dfrac{4}{5}=\boxed{-\dfrac{1}{5}}[/tex]