Упр.1400 ГДЗ Макарычев Миндюк 9 класс (Углубленный) (Алгебра)

а) sin(?)/(1-cos(?))+sin(?)/(1+cos(?)); в) sin(?)/(1+cos(?))+ctg(?);
б) cos(?)/(1+sin(?))+cos(?)/(1-sin(?)); г) cos(?)/(1-sin(?))-tg(?).

1. а)

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

   $$=\frac{\sin \beta+\sin \beta\cos \beta+\sin \beta-\sin \beta\cos \beta}{1-\cos^2 \beta} =\frac{2\sin \beta}{\sin^2 \beta} =\frac{2}{\sin \beta}$$

2. б)

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

   $$=\frac{\cos \beta-\sin \beta\cos \beta+\cos \beta+\sin \beta\cos \beta}{1-\sin^2 \beta} =\frac{2\cos \beta}{\cos^2 \beta} =\frac{2}{\cos \beta}$$

3. в)

   $$\frac{\sin a}{1+\cos a}+\operatorname{ctg} a= \frac{\sin a}{1+\cos a}+\frac{\cos a}{\sin a}$$

   $$=\frac{\sin^2 a+\cos a(1+\cos a)}{\sin a(1+\cos a)} =\frac{\sin^2 a+\cos a+\cos^2 a}{\sin a(1+\cos a)}$$

   $$=\frac{1+\cos a}{\sin a(1+\cos a)} =\frac{1}{\sin a}$$

4. г)

   $$\frac{\cos a}{1-\sin a}-\operatorname{tg} a= \frac{\cos a}{1-\sin a}-\frac{\sin a}{\cos a}$$

   $$=\frac{\cos^2 a-\sin a(1-\sin a)}{\cos a(1-\sin a)} =\frac{\cos^2 a-\sin a+\sin^2 a}{\cos a(1-\sin a)}$$

   $$=\frac{1-\sin a}{\cos a(1-\sin a)} =\frac{1}{\cos a}$$

Ответ

а) $$\frac{2}{\sin \beta}$$; б) $$\frac{2}{\cos \beta}$$; в) $$\frac{1}{\sin a}$$; г) $$\frac{1}{\cos a}$$.