Упр.22.13 ГДЗ Мерзляк 10 класс Базовый уровень (Алгебра)

Задача

1) (tg(5?/4)+tg(?/2-a))^2+(ctg(9?/4)+ctg(?-a))^2=2/sin^2(a);
2) cos^2(?/3+a)/tg^2(?/6-a)+sin^2(?/3+a)tg^2(?/6-a)=1;
3) cos^4(a-?)/(cos^4(a-3?/2)+sin^4(a+3?/2)-1)=-(1/2)ctg^2(a).

Подробный ответ

$$\left(\tg\frac{5\pi}{4}+\tg\left(\frac{\pi}{2}-a\right)\right)^2+\left(\ctg\frac{9\pi}{4}+\ctg(\pi-a)\right)^2$$
$$=\left(\tg\left(\pi+\frac{\pi}{4}\right)+\ctg a\right)^2+\left(\ctg\left(2\pi+\frac{\pi}{4}\right)-\ctg a\right)^2$$
$$=\left(\tg\frac{\pi}{4}+\ctg a\right)^2+\left(\ctg\frac{\pi}{4}-\ctg a\right)^2$$
$$=(1+\ctg a)^2+(1-\ctg a)^2$$
$$=1+2\ctg a+\ctg^2 a+1-2\ctg a+\ctg^2 a$$
$$=2(1+\ctg^2 a)=\frac{2}{\sin^2 a}.$$
Тождество доказано.
$$\frac{\cos^2\left(\frac{\pi}{3}+a\right)}{\tg^2\left(\frac{\pi}{6}-a\right)}+\sin^2\left(\frac{\pi}{3}+a\right)\tg^2\left(\frac{\pi}{6}-a\right)$$
$$=\cos^2\left(\frac{\pi}{2}-\left(\frac{\pi}{6}-a\right)\right)\cdot\frac{1}{\tg^2\left(\frac{\pi}{6}-a\right)}+\sin^2\left(\frac{\pi}{2}-\left(\frac{\pi}{6}-a\right)\right)\tg^2\left(\frac{\pi}{6}-a\right)$$
$$=\sin^2\left(\frac{\pi}{6}-a\right)\cdot\frac{\sin^2\left(\frac{\pi}{6}-a\right)}{\cos^2\left(\frac{\pi}{6}-a\right)}+\cos^2\left(\frac{\pi}{6}-a\right)\cdot\frac{\sin^2\left(\frac{\pi}{6}-a\right)}{\cos^2\left(\frac{\pi}{6}-a\right)}$$
$$=\sin^2\left(\frac{\pi}{6}-a\right)+\cos^2\left(\frac{\pi}{6}-a\right)=1.$$
Тождество доказано.
$$\frac{\cos^4(a-\pi)}{\cos^4\left(a-\frac{3\pi}{2}\right)+\sin^4\left(a+\frac{3\pi}{2}\right)-1}$$
$$=\frac{\cos^4 a}{\sin^4 a+\cos^4 a-1}$$
$$=\frac{\cos^4 a}{(1-\cos^2 a)^2+\cos^4 a-1}$$
$$=\frac{\cos^4 a}{1-2\cos^2 a+\cos^4 a+\cos^4 a-1}$$
$$=\frac{\cos^4 a}{-2\cos^2 a(1-\cos^2 a)}$$
$$=\frac{\cos^4 a}{-2\cos^2 a\sin^2 a}=-\frac{\cos^2 a}{2\sin^2 a}=-\frac12\ctg^2 a.$$
Тождество доказано.