Ctg^2 x-ctg^2 y= (cos^2 x-cos^2 y)/(sin^2 x-sin^2 y) Доказать тождество

$ctg^2a-ctg^2\beta =\frac{cos^2a}{sin^2a}-\frac{cos^2\beta }{sin^2\beta }=\frac{cos^2a\cdot sin^2\beta-cos^2\beta \cdot sin^2a}{sin^2a\cdot sin^2\beta }\; ;\\\\\\cos^2a\cdot sin^2\beta -cos^2\beta \cdot sin^2a=\\=(cosa\cdot sin\beta -cos\beta \cdot sina)(cosa\cdot sin\beta +cos\beta \cdot sina)=\\=sin(\beta -a)\cdot sin(a+\beta )=\\=\frac{1}{2}\cdot [cos((\beta -a)-(a+ \beta ))-cos((\beta -a)+(a+\beta ))\, ]=\\=\frac{1}{2}\cdot (cos2a-cos2 \beta )=\frac{1}{2}\cdot [\, (1-2sin^2a-(1-2sin^2\beta )\, ]=\\=-sin^2a+sin^2\beta =-(1-cos^2a)+(1-cos^2\beta )=\\=cos^2a-cos^2\beta \\\\ctg^2a-ctg^2 \beta =\frac{cos^2a-cos^2 \beta }{sin^2a\cdot sin^2 \beta }$

$P.S.\; \; \; \; cos^2a\cdot sin^2 \beta -cos^2 \beta \cdot sin^2a=\\=cos^2a\cdot (1-cos^2 \beta )-cos^2 \beta \cdot (1-cos^2a)=\\=cos^2a-cos^2a\cdot cos^2 \beta -cos^2 \beta +cos^2 \beta \cdot cos^2a=\\=cos^2a-cos^2 \beta$