Question

Допоможіть будласка
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Answer 1

$\displaystyle\bf\\\frac{Sin(\alpha -\beta )\cdot Sin(\alpha +\beta )}{Sin^{2}\beta -Sin^{2}\alpha } =\frac{\frac{1}{2}[Cos(\alpha -\beta -\alpha -\beta )-Cos(\alpha -\beta +\alpha +\beta )] }{Sin^{2}\beta -Sin^{2}\alpha} =\\\\\\=\frac{\frac{1}{2}(Cos2\beta -Cos2\alpha ) }{Sin^{2}\beta -Sin^{2} \alpha} =\frac{\frac{1}{2}(1-2Sin^{2} \beta -1+2Sin^{2} \alpha ) }{Sin^{2}\beta -Sin^{2}\alpha } =$

$\displaystyle\bf\\=\frac{\frac{1}{2} \cdot 2\cdot(Sin^{2} \alpha -Sin^{2}\beta ) }{Sin^{2} \beta -Sin^{2}\alpha } =-\frac{Sin^{2}\beta -Sin^{2} \alpha }{Sin^{2} \beta -Sin^{2}\alpha }=-1\\\\\\-1=-1$

Что и требовалось доказать