Question

Помогите пожалуйста решить алгебру вариант Б2 номер 3

Answer 1

А)cos⁴a + 2sin²a - sin⁴a = 1
cos⁴a + 2sin²a - sin⁴a = sin²a + cos²a
cos⁴a - sin⁴a = cos²a + sin²a - 2sin²a
(cos²a - sin²a)(cos²a + sin²a)=cos²a - sin²a
(cos²a - sin²a)*1=cos²a - sin²a
cos²a - sin²a=cos²a - sin²a
Тождество доказано.
б)cos a /(1-sin a)  -  cos a /(1+sin a) = 2 tg a
(cos a(1+sin a)-cos a(1-sin a) ) / ( (1-sin a)(1+sin a) )=2tg a
(cos a + sin a * cos a - cos a + sin a * cos a) / ( 1-sin² a)=2 tg a
(2sin a * cos a) / cos² a=2tga
2sin a / cos a=2 tg a
2tga=2tga
Тождество доказано.

Answer 2

$cos^4(a)+2sin^2(a)-sin^4(a)=1\\-(sin^4(a)-cos^4(a))+2sin^2(a)=1\\a^2-b^2=(a-b)(a+b)\\-((sin^2(a)-cos^2(a))(sin^2(a)+cos^2(a)))+2sin^2(a)=1\\sin^2(a)+cos^2(a)=1\\-(-(cos^2(a)-sin^2(a)))+2sin^2(a)=1\\-(-cos(2a))+2sin^2(a)=1\\cos(2a)+2sin^2(a)=1\\cos(2a)=cos^2(a)-sin^2(a)\\cos^2(a)-sin^2(a)+2sin^2(a)=1\\cos^2(a)+sin^2(a)=1\\1=1$
$\frac{cos(a)}{1-sin(a)} - \frac{cos(a)}{1+sin(a)} =2tg(a)\\\frac{cos(a)}{1-sin(a)} - \frac{cos(a)}{1+sin(a)} =2tg(a)\\ \frac{(1+sin(a))cos(a)-(1-sin(a))cos(a)}{(1-sin(a))(1+sin(a))} =2tg(a)\\ \frac{cos(a)+sin(a)cos(a)-(cos(a)-sin(a)cos(a))}{1-sin^2(a)} =2tg(a)\\ \frac{cos(a)+sin(a)cos(a)-cos(a)+sin(a)cos(a)}{cos^2(a)} =2tg(a)\\sin(ts)=sin(t)cos(s)+cos(t)sin(s)\\ \frac{sin(2a)}{cos^2(a)} =2tg(a)\\sin(2a)=2sin(a)cos(a)\\ \frac{2sin(a)cos(a)}{cos^2(a)} =2tg(a)\\ \frac{2sin(a)}{cos(a)} =2tg(a)\\2tg(a)=2tg(a)$