Предмет: Алгебра, автор: dove7549

Помогите это решить , рррррпрреееешить

Приложения:

Ответы

Автор ответа: Miroslava227
0

1.

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

2.

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

3.

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

4.

4 - tg \alpha  \times ctg \alpha  = 4 - 1 = 3

5.

 \cos {}^{2} ( \beta )  -  \cos {}^{2} ( \beta )  \sin {}^{2} ( \beta )  =  \cos {}^{2} ( \beta ) (1 -  \sin {}^{2} ( \beta ))  =  \\  =  \cos {}^{2} ( \beta )   \times \cos {}^{2} ( \beta )   =   \cos {}^{4} ( \beta )

6.

 \sin {}^{4} ( \beta )  +  \sin {}^{2} ( \beta )  \cos {}^{2} ( \beta )  = \\  =   \sin {}^{2} ( \beta ) ( \sin {}^{2} ( \beta )   + \cos {}^{2} ( \beta ))  =  \\  =  \sin {}^{2} ( \beta )   \times 1 = \sin {}^{2} ( \beta )

7.

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

8.

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

9.

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

10.

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

11.

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

12.

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

13.

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

14.

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

15.

 \frac{1 +  {ctg}^{4}  \alpha }{ {ctg}^{2}  \alpha  +  {tg}^{2}  \alpha }  = (1 +  {ctg}^{4}  \alpha ) \times  \frac{1}{ {ctg}^{2} \alpha  +  \frac{1}{ {ctg}^{2} \alpha  }  }  =  \\  = (1 +  {ctg}^{4}  \alpha ) \times  \frac{ {ctg}^{2}  \alpha }{1 +  {ctg}^{4}  \alpha  }  =  {ctg}^{2}  \alpha

16.

 \frac{ \cos {}^{2} ( \alpha ) }{1 +  {tg}^{2} \alpha  }  -  \frac{ \sin {}^{2} ( \alpha ) }{1 +  {ctg}^{2}  \alpha }  =  \\  =  \frac{ \cos {}^{2} ( \alpha ) }{ \frac{1}{ \cos {}^{2} ( \alpha ) } }  -  \frac{ \sin {}^{2} ( \alpha ) }{ \frac{1}{ \sin {}^{2} ( \alpha ) } }  =  \\  =  \cos {}^{4} ( \alpha )  -  \sin {}^{4} ( \alpha )  =   \\ = ( \cos {}^{2} ( \alpha )  -  \sin {}^{2} ( \alpha ))(  \cos {}^{2} ( \alpha )  +  \sin {}^{2} ( \alpha ) ) =  \\  =  \cos(2 \alpha )

Похожие вопросы
Предмет: Русский язык, автор: юлька196
Предмет: Литература, автор: илья19681