Question

Помогите, дам" много" баллов

Answer 1

Ответ:

1)

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

2)

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

3)

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

4)

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

5)

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

6)

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

7)

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

8)

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