Помогите,пожалуйста, с подробным решением

$\frac{sin(\frac{3\pi }{2}+\alpha)+sin(2\pi+\alpha)}{2cos(-\alpha)*sin(-\alpha)+1}=\frac{-cos\alpha+sin\alpha}{-2cos\alpha*sin\alpha+1}=\frac{sin\alpha-cos\alpha}{1-2cos\alpha*sin\alpha}=$ $=\frac{sin\alpha-cos\alpha}{cos^2\alpha+sin^2\alpha-2cos\alpha*sin\alpha}=\frac{sin\alpha-cos\alpha}{(sin\alpha-cos\alpha)^2}=\frac{1}{sin\alpha-cos\alpha}$

P.S.

$cos(-\alpha)=cos\alpha$

$sin(-\alpha)=-sin\alpha$

$sin(2\pi +\alpha)=sin\alpha$

$sin(\frac{3\pi }{2}+\alpha)=-cos\alpha$

$\frac{1}{sin\alpha-cos\alpha}=\frac{\frac{\sqrt{2} }{2} }{\frac{\sqrt{2} }{2}(sin\alpha-cos\alpha)} =\frac{\frac{\sqrt{2} }{2} }{\frac{\sqrt{2} }{2}*sin\alpha-\frac{\sqrt{2} }{2} cos\alpha} =\frac{\frac{\sqrt{2} }{2} }{cos\frac{\pi }{4}*sin\alpha-sin\frac{\pi }{4} cos\alpha} =$ $=\frac{\frac{\sqrt{2}}{2}}{sin(\alpha-\frac{\pi }{4})}=\frac{\sqrt{2} }{2sin(\alpha-\frac{\pi }{4})}=\frac{1}{\sqrt{2}sin(\alpha-\frac{\pi }{4})}$