Упр.21.10 ГДЗ Мерзляк 10 класс Базовый уровень (Алгебра)

1) (sin(a+?)-sin(?)cos(a))/(sin(a-?)+sin(?)cos(a))=1;
2) (v2cos(a)-2sin(45°-a))/(2sin(60°+a)-v3cos(a))=v2;
3) (2sin(a)cos(?)-sin(a-?))/(cos(a)cos(?)-sin(a)sin(?))=tg(a+?).

1. $$\frac{\sin(a+\beta)-\sin\beta\cdot\cos a}{\sin(a-\beta)+\sin\beta\cdot\cos a}$$
   $$=\frac{\sin a\cos\beta+\cos a\sin\beta-\sin\beta\cos a}{\sin a\cos\beta-\cos a\sin\beta+\sin\beta\cos a}$$
   $$=\frac{\sin a\cos\beta}{\sin a\cos\beta}=1.$$

2. $$\frac{\sqrt2\cos a-2\sin(45^\circ-a)}{2\sin(60^\circ+a)-\sqrt3\cos a}$$
   $$=\frac{\sqrt2\cos a-2(\sin45^\circ\cos a-\cos45^\circ\sin a)}{2(\sin60^\circ\cos a+\cos60^\circ\sin a)-\sqrt3\cos a}$$
   $$=\frac{\sqrt2\cos a-2\left(\frac{\sqrt2}{2}\cos a-\frac{\sqrt2}{2}\sin a\right)}{2\left(\frac{\sqrt3}{2}\cos a+\frac12\sin a\right)-\sqrt3\cos a}$$
   $$=\frac{\sqrt2\cos a-\sqrt2\cos a+\sqrt2\sin a}{\sqrt3\cos a+\sin a-\sqrt3\cos a}$$
   $$=\frac{\sqrt2\sin a}{\sin a}=\sqrt2.$$

3. $$\frac{2\sin a\cdot\cos\beta-\sin(a-\beta)}{\cos a\cdot\cos\beta-\sin a\cdot\sin\beta}$$
   $$=\frac{2\sin a\cos\beta-(\sin a\cos\beta-\cos a\sin\beta)}{\cos(a+\beta)}$$
   $$=\frac{\sin a\cos\beta+\cos a\sin\beta}{\cos(a+\beta)}$$
   $$=\frac{\sin(a+\beta)}{\cos(a+\beta)}=\tg(a+\beta).$$

Ответ

$$1) \ 1;\qquad 2) \ \sqrt2;\qquad 3) \ \tg(a+\beta).$$