Question

= ДАЮ 100 БАЛЛОВ ЗА ПРАВИЛЬНОЕ РЕШЕНИЕ (1,2,3,6) =

Answer 1

Ответ:

1.

$3tg( - \frac{\pi}{6} )ctg \frac{\pi}{3} + \sin( \frac{3\pi}{2} ) - 4 \cos( \frac{\pi}{4} ) = \\ = 3 \times ( - \frac{ \sqrt{3} }{3} ) \times \frac{ \sqrt{3} }{3} - 1 - 4 \times \frac{ \sqrt{2} }{2} = \\ = - 1 - 1 - 2 \sqrt{2} = - 2 - 2 \sqrt{2}$

2.

1)

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

2)

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

3.

1)

$\cos(156^{\circ}) < 0 \\ \sin( - 350^{\circ}) > 0 \\ ctg(230^{\circ}) > 0 \\ \\ = > \cos(156^{\circ}) \sin( - 350^{\circ}) ctg(230^{\circ}) < 0$

2)

$\cos( \frac{13\pi}{15} ) < 0 \\ ctg( \frac{23\pi}{18} ) > 0 \\ \\ = > \cos( \frac{13\pi}{15} ) ctg( \frac{23\pi}{18} ) < 0$

6.

1)

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

2)

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