Математика: сделать четные номера,сколько сможете

сделать четные номера,сколько сможете

Ответ:

оллл11

19.05.2021 15:12

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

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

$(tg \alpha + ctg \alpha ) {}^{2} - (tg \alpha - ctg \alpha ) {}^{2} = \\ = (tg \alpha + ctg \alpha - tg \alpha + ctg \alpha )(tg \alpha + ctg \alpha + tg \alpha - ctg \alpha ) = \\ = 2ctg \alpha \times 2tg \alpha = 4$

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

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

$\sin {}^{4} ( \alpha ) + \sin {}^{2} ( \alpha ) \cos {}^{2} ( \alpha ) + \cos {}^{2} ( \alpha ) = \\ = \sin {}^{2} ( \alpha ) ( \sin {}^{2} ( \alpha ) + \cos {}^{2} ( \alpha ) ) + \cos {}^{2} ( \alpha ) = \\ = \sin {}^{2} ( \alpha ) + \cos {}^{2} ( \alpha ) = 1$