Question

алгебра, срочно......​

Answer 1

$\displaystyle\bf\\\frac{Sin\Big(\dfrac{3\pi }{2} +\alpha \Big)+Sin\Big(2\pi+ \alpha \Big)}{2Cos\Big(-\alpha \Big) Sin\Big(-\alpha \Big)+1} =\frac{-Cos\alpha+Sin\alpha }{2Cos\alpha \cdot\Big(-Sin\alpha \Big)+1} =\\\\\\=\frac{Sin\alpha -Cos\alpha }{1-2Sin\alpha Cos\alpha }=\frac{Sin\alpha -Cos\alpha }{Sin^{2}\alpha -2Sin\alpha Cos\alpha+Cos^{2} \alpha } =\\\\\\=\frac{Sin\alpha -Cos\alpha }{(Sin\alpha -Cos\alpha )^{2} } =\frac{1}{Sin\alpha -Cos\alpha }$