Question

ПОМОГИТЕ ПОЖАЛУЙСТА, 2 и 4​

Answer 1

Ответ:

2.

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

4.

$\frac{ { \sin }^{2} (3\pi - \alpha )}{1 - \cos( - \alpha ) } + \cos(5\pi - \alpha ) = \\ = \frac{ { \sin}^{2} \alpha }{1 - \cos( \alpha ) } - \cos( \alpha ) = \\ = \frac{1 - { \cos }^{2} \alpha }{ 1 - \cos( \alpha ) } - \cos( \alpha ) = \\ = \frac{(1 - \cos( \alpha ))(1 + \cos( \alpha) ) }{1 - \cos( \alpha ) } - \cos( \alpha ) = \\ = 1 + \cos( \alpha ) - \cos( \alpha ) = 1$

Answer 2

$2)\frac{Sin^{2}(\pi+\alpha)}{1-Cos\alpha}-Cos(2\pi-\alpha)=\frac{Sin^{2}\alpha}{1-Cos\alpha}-Cos\alpha=\frac{Sin^{2}\alpha-Cos\alpha+Cos^{2}\alpha }{1-Cos\alpha }=\\\\=\frac{\underbrace{Sin^{2}\alpha+Cos^{2}\alpha}_{1}-Cos\alpha}{1-Cos\alpha}=\frac{1-Cos\alpha }{1-Cos\alpha} =\boxed1$

$4)\frac{Sin^{2}(3\pi-\alpha)}{1-Cos(-\alpha)}+Cos(5\pi-\alpha)=\frac{Sin^{2} \alpha }{1-Cos\alpha}-Cos\alpha=\frac{Sin^{2}\alpha+Cos^{2}\alpha-Cos\alpha}{1-Cos\alpha}=\\\\=\frac{1-Cos\alpha }{1-Cos\alpha }=\boxed1$