Предмет: Алгебра, автор: galekkiler

Помогите пожалуйста дам 80 балл. Алгебра:) ​

Приложения:

Ответы

Автор ответа: Miroslava227
2

22.5

1.

 \frac{ \sin(x) +  \cos(x)  }{1 + tgx}  =  \frac{ \sin(x)  + \cos(x)  }{1 +  \frac{ \sin(x) }{ \cos(x) } }  =  \\  = ( \sin(x)  +  \cos(x))  \times  \frac{ \cos(x) }{ \sin(x)  + \cos(x)  }  =  \\  =  \cos(x)

2.

 \frac{ctgx - 1}{ \sin(x) - \cos(x)  }  =  \frac{ \frac{ \cos(x) }{ \sin(x) } - 1 }{ \sin(x) -  \cos(x)  }  =  \\  =  \frac{ \cos(x)  - \sin(x)  }{ \sin(x) }  \times  \frac{1}{ \sin(x) -  \cos(x)  }  =  \\  =  \frac{ - ( \sin(x) -  \cos(x))  }{ \sin(x) \times ( \sin(x)   -  \cos(x)) }  =  \\  =  -  \frac{1}{ \sin(x) }

3.

 \frac{1 + ctgx}{ \sin(x)  +  \cos(x) }  =  \frac{1 +  \frac{ \cos(x) }{ \sin(x) } }{ \sin(x)   + \cos(x) }  =  \\  =  \frac{ \sin(x)  +  \cos(x) }{ \sin(x) }  \times  \frac{1}{ \sin(x)  + \cos(x)  }  =  \\  =  \frac{1}{ \sin(x) }

4.

 \frac{ \sin(x)  -  \cos(x) }{1 - tgx}  =  \frac{ \sin(x) -  \cos(x)  }{1 -  \frac{ \sin(x) }{ \cos(x) } }  =  \\  = ( \sin(x) -   \ cos(x) ) \times  \frac{ \cos(x) }{ \cos(x)  -  \sin(x) }  =  \\  =  \frac{ - ( \sin(x)   - \cos(x) ) \times  \cos(x) }{ \sin(x)   - \cos(x) }  =  -  \cos(x)

22.6

1.

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

2.

 { \sin }^{2}  \gamma  +  \frac{ {ctg}^{2}  \gamma  - 1}{ {ctg}^{2} \gamma  + 1 }  =  { \sin }^{2}  \gamma  +  \frac{ \frac{ { \cos}^{2} \gamma  }{ { \sin}^{2}  \gamma } - 1 }{ \frac{ { \cos}^{2}  \gamma }{ { \sin }^{2} \gamma  } + 1 }  =  \\  =  { \sin }^{2}  \gamma  +  \frac{ { \cos }^{2}  \gamma  -  { \sin}^{2}  \gamma }{ {  \sin }^{2} \gamma  }  \times  \frac{ { \sin}^{2} \gamma    }{ {  \sin  }^{2} \gamma  +  { \cos }^{2}  \gamma  }  =  \\  =  { \sin}^{2}  \gamma  +  { \cos }^{2}  \gamma  -  { \sin }^{2}  \gamma  =  { \cos }^{2}  \gamma

3.

 \frac{ {ctg}^{2}  \gamma  - 1}{ {ctg}^{2} \gamma  + 1 }  -  { \cos }^{2}  \gamma  =  \frac{ \frac{ { \cos }^{2} \gamma  }{ { \sin}^{2} \gamma  }  - 1}{  \frac{ { \cos }^{2}  \gamma }{ { \sin }^{2} \gamma  }  + 1 }  -  { \cos}^{2}  \gamma  =  \\  =  \frac{ { \cos }^{2} \gamma  -  { \sin }^{2} \gamma   }{ { \sin }^{2}  \gamma } \times  \frac{ { \sin }^{2} \gamma  }{ { \cos}^{2}  \gamma  +  { \sin }^{2}  \gamma }   -  { \cos }^{2}  \gamma  =  \\  = { \cos }^{2}   \gamma  -  { \sin }^{2}  \gamma  -  { \cos }^{2}  \gamma  =  \\  =  -  { \sin }^{2}  \gamma

4.

 \frac{ {tg}^{2}x - 1 }{ {tg}^{2}x + 1 }  -  { \sin}^{2} x =  \frac{ \frac{ { \sin}^{2}x }{ { \cos }^{2}x }  - 1}{  \frac{ { \sin }^{2}x }{ { \cos }^{2} x}    + 1}  -  { \sin}^{2} x =  \\  =  \frac{ { \sin }^{2}x -  { \cos }^{2}x  }{ { \cos }^{2}x }  \times  \frac{ { \cos }^{2} x}{ { \sin }^{2}x +  { \cos }^{2} x }  -  { \sin }^{2} x =  \\  =  { \sin }^{2} x -  { \cos }^{2} x -  { \sin }^{2} x =  -  { \cos }^{2} x


galekkiler: Спасибо ♡
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