Question

ПОЖАЛУЙСТА ПОЛНОСТЬЮ РЕШИТЕ 2 ВАРИАНТ, 50 БАЛЛОВ!!!!​

Answer 1

Ответ:

$a) \cos^{2} \beta - cos^{2} \beta \times \sin^{2} \beta = \cos^{2} \beta (1 - \sin ^{2} \beta ) = cos^{2} \beta \times cos^{2} \beta = cos^{4} \beta \\ b) tg^{2} \beta \times ctg^{2} \beta - cos^{2} \beta = \frac{sin^{2} \beta}{cos^{2} \beta} \times \frac{cos^{2} \beta}{sin^{2} \beta} - = 1 - \cos^{2} \beta = sin^{2} \beta \\ \frac{sin^{2} \beta - cos^{2} \beta + 1}{sin^{2} \beta} = \frac{1}{sin^{2} \beta }$

sin²A-cos²A+1=1-cos²A-cos²A+cos²A+sin²A=1-cos²A+sin²A=sin²A+cos²A-cos²A+sin²A=2sin²A

2sin²A/sin²A=2

sinA=√1-cos²A=√1-64/289=√225/289=15/17

a{8/17:5/17}