Question

​буду !

Answer 1

$\frac{ \tan\alpha \cos \alpha }{ 1 + { \cot}^{2} \alpha } = \frac{ \frac{ \sin\alpha }{ \cos\alpha } \cos \alpha }{ 1 + \frac{ { \cos }^{2} \alpha }{ { \sin}^{2} \alpha } } = \\ = \frac{ \sin\alpha }{ \frac{ { \sin}^{2} \alpha + { \cos }^{2} \alpha }{ { \sin}^{2} \alpha } } = \frac{ \sin \alpha }{ \frac{1}{ { \sin }^{2} \alpha } } = { \sin }^{3} \alpha$