[tex]\displaystyle\bf\\\frac{Sin3\alpha -Sin\alpha }{Cos3\alpha +Cos\alpha } =\frac{2Sin\dfrac{3\alpha -\alpha }{2} Cos\dfrac{3\alpha +\alpha }{2} }{2Cos\dfrac{3\alpha -\alpha }{2} Cos\dfrac{3\alpha +\alpha }{2} } =\frac{2Sin\alpha Cos2\alpha }{2Cos\alpha Cos2\alpha } =\\\\\\=\frac{Sin\alpha }{Cos\alpha } =tg\alpha \\\\\\tg\alpha =tg\alpha[/tex]

Тлождество доказано