Verified answer

[tex]\displaystyle\bf\\\frac{Sin\Big(\pi -\alpha \Big)+Cos\Big(\dfrac{\pi }{2}-\alpha \Big) }{Sin\Big(\dfrac{3\pi }{2} -\alpha \Big)} =\frac{Sin\alpha +Sin\alpha }{-Cos\alpha }=-\frac{2Sin\alpha }{Cos\alpha } =-2tg\alpha[/tex]