Question

$\frac{2Cos(\pi-\beta)-Sin(-\frac{\pi} {2}+ \beta)}{Cos(\beta-\pi) }$

Пожалуйста!!!

Answer 1

$\frac{2Cos(\pi-\beta)-Sin(-\frac{\pi} {2}+ \beta)}{Cos(\beta-\pi) } = \\ = \frac{2 \sin( \beta ) - ( - \cos( \beta )) }{ - \cos( \beta ) } = \\ = - 2 \frac{ \sin( \beta ) }{ \cos( \beta ) } - \frac{ \cos( \beta ) }{ \cos( \beta ) } = \\ = - 2 \tan( \beta ) - 1$