Question

Помогите с матешей :^​

Answer 1

$\dfrac{\sin^2\alpha}{1+\cos\alpha}+\cos \alpha =\dfrac{(1-\cos \alpha)(1+\cos \alpha )}{1+\cos \alpha}+\cos \alpha=1-\cos \alpha +\cos \alpha =1$

$\dfrac{2\sin^2\alpha \cdot {\rm ctg}\, \alpha}{\cos^2\alpha-\sin^2\alpha}=\dfrac{2\sin \alpha\cos \alpha}{\cos2\alpha}=\dfrac{\sin2\alpha}{\cos 2\alpha}={\rm tg}\, 2\alpha$