Założenie:

$\alpha \ne k\cdot {90}^{\circ },k\in \mathbb{Z}$  

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Dowód:

Zauważmy, że:

${\cos }^{}\left(270^{\circ }-\alpha \right)=-\sin \alpha$

${{\cos }^2\left({}^{}270^{\circ }-\alpha \right)=\left(-\sin \alpha \right)}^2={\sin }^2\alpha$

a także

$\sin \left(270^{\circ }+\alpha \right)=-\cos \alpha$

${\sin }^2\left(270^{\circ }+\alpha \right)={\cos }^2\alpha$

oraz

$\sin \left(90^{\circ }+\alpha \right)=\cos \alpha$

$\sin {}^2\left(90^{\circ }+\alpha \right)={\cos }^2\alpha$

i

$\cos \left(90^{\circ }-\alpha \right)=\sin \alpha$

$\cos {}^2\left(90^{\circ }-\alpha \right)={\sin }^2\alpha$

Zatem:

$\frac{{\cos }^2\left({270}^{\circ }-\alpha \right)}{{\sin }^{-2}\left(\alpha +{90}^{\circ }\right)-1}+\frac{{\sin }^2\left(\alpha +{270}^{\circ }\right)}{{\cos }^{-2}\left({90}^{\circ }-\alpha \right)-1}=$  

$=\frac{{\sin }^2\alpha }{\frac{1}{{\sin }^2\left(\alpha +90^{\circ }\right)}-1}+\frac{{\cos }^2\alpha }{\frac{1}{{\cos }^2\left(90^{\circ }-\alpha \right)}-1}=$

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

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

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

A to należało wykazać.