Question

СРОЧНО! ДАЮ 60 БАЛОВ.

Спростить г.

Answer 1

Ответ:

а

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

б

$\sin( \alpha ) \sin( 180^{\circ} - \alpha ) - \cos( \alpha ) \cos( 180^{\circ} - \alpha ) = \\ = \sin( \alpha ) \times \sin( \alpha ) + \cos( \alpha ) \times \cos( \alpha ) = 1$

в

$\sin {}^{2} ( 180 ^{\circ} - \alpha ) + \cos {}^{2} ( \alpha ) + \frac{ \sin {}^{2} ( \alpha ) }{ \cos {}^{2} ( 180^{\circ} - \alpha ) } = \\ = \sin{}^{2} ( \alpha ) + \cos {}^{2} ( \alpha ) + \frac{ \sin {}^{2} ( \alpha ) }{ \cos {}^{2} ( \alpha ) } = 1 + {tg}^{2} \alpha = \frac{1}{ \cos {}^{2} ( \alpha ) }$

г

$\frac{ \cos( 180 ^{\circ} - \alpha ) }{ \cos( 90^{\circ} - \alpha ) } \times \frac{ \sin( \alpha ) }{ \sin( 90 ^{\circ} - \alpha ) } = \frac{ - \cos( \alpha ) }{ \sin( \alpha ) } \times \frac{ \sin( \alpha ) }{ \cos( \alpha ) } = - 1$