Question

помогите пожалуйста. Первый и второй вариант

Answer 1

Ответ:

1 вариант

1.

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

2.

$5 \sin(60 ^{\circ} ) - 4tg(45^{\circ} ) \cos(30^{\circ} ) = \\ = 5 \times \frac{ \sqrt{3} }{2} - 4 \times 1 \times \frac{ \sqrt{3} }{2} = \\ = \frac{5 \sqrt{3} - 4 \sqrt{3} }{2} = \frac{ \sqrt{3} }{2}$

3.

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

2 вариант

1.

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

2.

$3tg(30^{\circ} ) - 2 \sqrt{2} \sin(60^{\circ} ) \cos(45^{\circ} ) = \\ = 3 \times \frac{ \sqrt{3} }{3} - 2 \sqrt{2} \times \frac{ \sqrt{3} }{2} \times \frac{ \sqrt{2} }{2} = \\ = \sqrt{3} - \frac{ \sqrt{3} }{2} = \frac{ \sqrt{3} }{2}$

3.

$\cos( \alpha ) = \sqrt{1 - \sin {}^{2} ( \alpha ) } = \\ = \sqrt{1 - \frac{225}{289} } = \sqrt{ \frac{64}{289} } = \frac{8}{17}$