Question

упростите выражение
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Answer 1

$\displaystyle\bf\\\Big(\frac{Sin\alpha }{1+Cos\alpha } +\frac{Sin\alpha }{1-Cos\alpha } \Big)\cdot\frac{1}{1+Ctg^{2} \alpha }=\\\\\\\frac{Sin\alpha -Sin\alpha Cos\alpha +Sin\alpha +Sin\alpha Cos\alpha }{(1+Cos\alpha )(1-Cos\alpha )} \cdot\frac{1}{\frac{1}{Sin^{2}\alpha } } =\\\\\\=\frac{2Sin\alpha }{1-Cos^{2} \alpha } \cdot Sin^{2} \alpha =\frac{2Sin\alpha \cdot Sin^{2} \alpha }{Sin^{2}\alpha } =2Sin\alpha$