Question

ПОЖАЛУЙСТА ПОМОГИТЕ С АЛГЕБРОЙ!!!!
9 КЛАСС
СРОЧНО!!!!
даю 100б​

Answer 1

Ответ:

Объяснение:

sin²a/sin(a-b)+sin²b/sin(b-a)=sin²a/sin(a-b)-sin²b/sin(a-b)=

=(sin²a-sin²b)/sin(a-b)=

=2*sin((a+b)/2)*cos((a+b)/2)*2*sin((a-b)/2)*cos((a-b)/2)/sin(a-b)=

=sin(a+b)*sin(a-b)/sin(a-b)=sin(a+b)

Answer 2

$\displaystyle\bf\\\frac{Sin^{2}\alpha }{Sin(\alpha -\beta )} +\frac{Sin^{2} \beta }{Sin(\beta -\alpha )} =\frac{Sin^{2}\alpha }{Sin(\alpha -\beta )} -\frac{Sin^{2} \beta }{Sin(\alpha-\beta )} =\\\\\\=\frac{Sin^{2} \alpha -Sin^{2} \beta }{Sin(\alpha -\beta )} =\frac{(Sin\alpha -Sin\beta )\cdot(Sin\alpha +Sin\beta )}{Sin(\alpha -\beta )} =\\\\\\=\frac{2Sin\dfrac{\alpha -\beta }{2} Cos\dfrac{\alpha +\beta }{2}\cdot 2Sin\dfrac{\alpha +\beta }{2} Cos\dfrac{\alpha -\beta }{2} }{Sin(\alpha- \beta )} =$

$\displaystyle\bf\\=\frac{\Big(2Sin\dfrac{\alpha -\beta }{2} Cos\dfrac{\alpha -\beta }{2}\Big)\cdot \Big(2Sin\dfrac{\alpha +\beta }{2} Cos\dfrac{\alpha +\beta }{2} \Big) }{Sin(\alpha- \beta )} =\\\\\\=\frac{Sin(\alpha -\beta )\cdot Sin(\alpha +\beta )}{Sin(\alpha -\beta )} =Sin(\alpha +\beta )$