Упр.20.11 ГДЗ Мерзляк 10 класс Базовый уровень (Алгебра)

Задача

1) (cos^3(a)-sin^3(a))/(1+sin(a)cos(a))=cos(a)-sin(a);
2) tg^2(a)-sin^2(a)=tg^2(a)sin^2(a);
3) (v3-2sin(a))/(2cos(a)-1)=(1+2cos(a))/(2sin(a)+v3);
4) (ctg^2(a)-cos^2(a))/(sin^2(a)-tg^2(a))=-ctg^6(a);
5) (sin(a)+tg(a))/(1+cos(a))=tg(a);
6) (sin^2(a))/(ctg^2(a)-cos^2(a))=tg^4(a).

Подробный ответ

$$\frac{\cos^3 a-\sin^3 a}{1+\sin a\cos a}=\frac{(\cos a-\sin a)(\cos^2 a+\sin a\cos a+\sin^2 a)}{1+\sin a\cos a}$$
$$=\frac{(\cos a-\sin a)(1+\sin a\cos a)}{1+\sin a\cos a}=\cos a-\sin a.$$
$$\tg^2 a-\sin^2 a=\frac{\sin^2 a}{\cos^2 a}-\sin^2 a$$
$$=\sin^2 a\left(\frac{1}{\cos^2 a}-1\right)=\sin^2 a\cdot\frac{1-\cos^2 a}{\cos^2 a}$$
$$=\sin^2 a\cdot\frac{\sin^2 a}{\cos^2 a}=\tg^2 a\cdot\sin^2 a.$$
$$\frac{\sqrt3-2\sin a}{2\cos a-1}=\frac{1+2\cos a}{2\sin a+\sqrt3}$$
$$\frac{(\sqrt3-2\sin a)(2\sin a+\sqrt3)-(1+2\cos a)(2\cos a-1)}{(2\cos a-1)(2\sin a+\sqrt3)}=0.$$
Раскроем скобки:
$$3-4\sin^2 a-4\cos^2 a+1=0,$$
$$4-4(\sin^2 a+\cos^2 a)=0,$$
$$4-4=0.$$
Следовательно, тождество верно.
$$\frac{\ctg^2 a-\cos^2 a}{\sin^2 a-\tg^2 a}=\frac{\frac{\cos^2 a}{\sin^2 a}-\cos^2 a}{\sin^2 a-\frac{\sin^2 a}{\cos^2 a}}$$
$$=\frac{\cos^2 a\left(\frac{1}{\sin^2 a}-1\right)}{\sin^2 a\left(1-\frac{1}{\cos^2 a}\right)}$$
$$=\frac{\cos^2 a\cdot\frac{1-\sin^2 a}{\sin^2 a}}{\sin^2 a\cdot\frac{\cos^2 a-1}{\cos^2 a}}$$
$$=\frac{\cos^2 a\cdot\frac{\cos^2 a}{\sin^2 a}}{\sin^2 a\cdot\frac{-\sin^2 a}{\cos^2 a}}=-\frac{\cos^6 a}{\sin^6 a}=-\ctg^6 a.$$
$$\frac{\sin a+\tg a}{1+\cos a}=\frac{\sin a+\frac{\sin a}{\cos a}}{1+\cos a}$$
$$=\frac{\sin a\cos a+\sin a}{\cos a(1+\cos a)}=\frac{\sin a(\cos a+1)}{\cos a(1+\cos a)}$$
$$=\frac{\sin a}{\cos a}=\tg a.$$
$$\frac{\sin^2 a}{\ctg^2 a-\cos^2 a}=\frac{\sin^2 a}{\frac{\cos^2 a}{\sin^2 a}-\cos^2 a}$$
$$=\frac{\sin^2 a}{\cos^2 a\left(\frac{1}{\sin^2 a}-1\right)}=\frac{\sin^2 a}{\cos^2 a\cdot\frac{1-\sin^2 a}{\sin^2 a}}$$
$$=\frac{\sin^2 a}{\cos^2 a\cdot\frac{\cos^2 a}{\sin^2 a}}=\frac{\sin^4 a}{\cos^4 a}=\tg^4 a.$$