№22.4 пж, пж, пж Прям щас нужно. Created by serikbekadilhan3. algebra-ru

№22.4 пж, пж, пж Прям щас нужно...

serikbekadilhan3 @serikbekadilhan3

July 2022 1 48 Report

№22.4 пж, пж, пж
Прям щас нужно

Answers & Comments

Miroslava227

Ответ:

$\frac{ \sin( \alpha ) \cos( \alpha ) }{tg \alpha } - 1 = \sin( \alpha ) \cos( \beta ) \times \frac{ \cos( \alpha ) }{ \sin( \alpha ) } - 1 = \\ = { \cos}^{2} \alpha - 1 = - { \sin }^{2} \alpha$

$\frac{ \sin( \alpha ) \cos( \alpha ) }{ctg \alpha } - 1 = \sin( \alpha ) \cos( \alpha ) \times \frac{ \sin( \alpha ) }{ \cos( \alpha ) } - 1 = \\ = { \sin }^{2} \alpha - 1 = - { \cos }^{2} \alpha$

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

$\frac{1 + tg \alpha }{1 + ctg \alpha } = \frac{1 + \frac{ \sin( \alpha ) }{ \cos( \alpha ) } }{1 + \frac{ \cos( \alpha ) }{ \sin( \alpha ) } } = \\ = \frac{ \cos( \alpha ) + \sin( \alpha ) }{ \cos( \alpha ) } \times \frac{ \sin( \alpha ) }{ \sin( \alpha ) + \cos( \alpha ) } = \\ = \frac{ \sin( \alpha ) }{ \cos( \alpha ) } = tg \alpha$

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