Алгебра: Алгебра 10клас ! Задания на фото

Алгебра 10клас ! Задания на фото

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Ответ:

matveq888

03.04.2021 09:27

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

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

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

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

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

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Ответ:

3462686

03.04.2021 09:27

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

$\frac{1-\cos^2{\alpha}}{1-\sin^2{\alpha}}+tg \, \alpha \cdot ctg \, \alpha=\frac{\sin^2{\alpha}}{\cos^2{\alpha}}+1=tg^2\, \alpha +1 = \frac{1}{\cos^2{\alpha}}$

$\frac{1}{1+tg^2 \, \alpha}+\sin^2{\alpha}=\frac{1}{\frac{1}{\cos^2{\alpha}}}+\sin^2{\alpha}=\cos^2{\alpha}+\sin^2{\alpha}=1$

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

$=-\frac{1-\cos{\alpha}}{\sin{\alpha}\cdot (1-\cos{\alpha)}}=-\frac{1}{\sin{\alpha}}$

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