Question

помогите пожалуйста

Answer 1

$\displaystyle\bf\\\frac{Sin(\alpha +40^\circ)-Cos\alpha \cdot Sin40^\circ}{Cos\alpha \cdot Cos40^\circ} =\\\\\\=\frac{Sin\alpha\cdot Cos40^\circ+Cos\alpha\cdot Sin40^\circ-Cos\alpha \cdot Sin40^\circ}{Cos\alpha \cdot Cos40^\circ} =\\\\\\=\frac{Sin\alpha \cdot Cos40^\circ}{Cos\alpha \cdot Cos40^\circ} =\frac{Sin\alpha }{Cos\alpha } =tg\alpha$