Question

Доведіть тотожність
cos(-a) - tg (pi/2+ a) / 1- sin(2pi - a) =ctg a

Answer 1

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

Тождество доказано