Question

Помогите пожалуйста, развёрнутый ответ

Answer 1

Ответ:

1.

угол принадлежит 4 четверти, cosa > 0

$\sin( \alpha ) = - \frac{5}{13} \\ \cos( \alpha ) = \sqrt{1 - \sin {}^{2} ( \alpha ) } \\ \cos( \alpha ) = \sqrt{1 - \frac{25}{169} } = \sqrt{ \frac{144}{169} } = \frac{12}{13}$

$tg \alpha = \frac{ \sin( \alpha ) }{ \cos( \alpha ) } = - \frac{5}{13} \times \frac{13}{12} = - \frac{5}{12} \\ ctg \alpha = \frac{1}{tg \alpha } = - \frac{12}{5}$

2.

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

3.

$\frac{1 - \cos {}^{2} (\phi) }{1 - \sin {}^{2} (\phi) } + tg \frac{2}{3} ctg \frac{2}{3} = \\ = \frac{ \sin {}^{2} (\phi) }{ \cos {}^{2} ( \phi ) } + tg \frac{2}{3} \times \frac{1}{tg \frac{2}{3} } = \\ = {tg}^{2} (\phi) + 1 = \frac{1}{ \cos {}^{2} (\phi) }$

4.

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