Question

Даю 50 баллов срочно надо чтобы сделали ​

Answer 1

$\displaystyle\bf\\\frac{Sin20^\circ Cos10^\circ+Cos160^\circ Cos100^\circ}{Sin21^\circ Cos9^\circ+Cos159^\circ Cos99^\circ} =\\\\\\=\frac{Sin20^\circ Cos10^\circ+Cos(180^\circ-20 ^\circ )Cos(90^\circ+10^\circ) }{Sin21^\circ Cos9^\circ+Cos(180^\circ-21^\circ) Cos(90^\circ+9^\circ) } =\\\\\\=\frac{Sin20^\circ Cos10^\circ+Cos20^\circ Sin10^\circ}{Sin21^\circ Cos9^\circ+Cos21^\circ Sin9^\circ} =\frac{Sin(20^\circ+10^\circ) }{Sin(21^\circ+9^\circ) } =\\\\\\=\frac{Sin30^\circ}{Sin30^\circ} =1$