Упр.25.7 ГДЗ Мерзляк 10 класс Углубленный уровень (Алгебра)

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

1. $$\frac{\cos^2\left(\frac{\pi}{3}+\alpha\right)}{\tg^2\left(\frac{\pi}{6}-\alpha\right)}+\sin^2\left(\frac{\pi}{3}+\alpha\right)\tg^2\left(\frac{\pi}{6}-\alpha\right)$$
   $$=\cos^2\left(\frac{\pi}{3}+\alpha\right)\ctg^2\left(\frac{\pi}{6}-\alpha\right)+\sin^2\left(\frac{\pi}{3}+\alpha\right)\tg^2\left(\frac{\pi}{6}-\alpha\right)$$
   $$=\cos^2\left(\frac{\pi}{2}-\left(\frac{\pi}{6}-\alpha\right)\right)\ctg^2\left(\frac{\pi}{6}-\alpha\right)+\sin^2\left(\frac{\pi}{2}-\left(\frac{\pi}{6}-\alpha\right)\right)\tg^2\left(\frac{\pi}{6}-\alpha\right)$$
   $$=\sin^2\left(\frac{\pi}{6}-\alpha\right)\ctg^2\left(\frac{\pi}{6}-\alpha\right)+\cos^2\left(\frac{\pi}{6}-\alpha\right)\tg^2\left(\frac{\pi}{6}-\alpha\right)$$
   $$=\sin^2\left(\frac{\pi}{6}-\alpha\right)+\cos^2\left(\frac{\pi}{6}-\alpha\right)=1.$$
2. $$\frac{\cos^4(\alpha-\pi)}{\cos^4\left(\alpha-\frac{3\pi}{2}\right)+\sin^4\left(\alpha+\frac{3\pi}{2}\right)-1}$$
   $$=\frac{\cos^4\alpha}{\sin^4\alpha+\cos^4\alpha-\sin^2\alpha-\cos^2\alpha}$$
   $$=\frac{\cos^4\alpha}{\sin^4\alpha+\cos^4\alpha-1}$$
   $$=\frac{\cos^4\alpha}{\sin^4\alpha+\cos^4\alpha-(\sin^2\alpha+\cos^2\alpha)}$$
   $$=\frac{\cos^4\alpha}{\sin^4\alpha+\cos^4\alpha-\sin^2\alpha-\cos^2\alpha}$$
   $$=\frac{\cos^4\alpha}{\sin^2\alpha(\sin^2\alpha-1)+\cos^2\alpha(\cos^2\alpha-1)}$$
   $$=\frac{\cos^4\alpha}{-\sin^2\alpha\cos^2\alpha-\cos^2\alpha\sin^2\alpha}$$
   $$=\frac{\cos^4\alpha}{-2\sin^2\alpha\cos^2\alpha}=-\frac{1}{2}\ctg^2\alpha.$$
3. $$\sin^2\left(\frac{3\pi}{2}-\alpha\right)(\tg^2\alpha-1)\ctg\left(\alpha-\frac{5\pi}{4}\right)\sin^{-2}\left(\frac{5\pi}{4}+\alpha\right)$$
   $$=\cos^2\alpha(\tg^2\alpha-1)\ctg\left(\alpha-\frac{5\pi}{4}\right)\sin^{-2}\left(\frac{5\pi}{4}+\alpha\right)$$
   $$=(\sin^2\alpha-\cos^2\alpha)\ctg\left(\alpha-\frac{5\pi}{4}\right)\sin^{-2}\left(\frac{5\pi}{4}+\alpha\right)$$
   $$=-\cos 2\alpha\cdot \ctg\left(\alpha-\frac{5\pi}{4}\right)\cdot \sin^{-2}\left(\alpha+\frac{5\pi}{4}\right).$$
   Так как
   $$\ctg\left(\alpha-\frac{5\pi}{4}\right)=\ctg\left(\alpha+\frac{\pi}{4}\right),$$
   $$\sin\left(\alpha+\frac{5\pi}{4}\right)=-\sin\left(\alpha+\frac{\pi}{4}\right),$$
   то
   $$-\cos 2\alpha\cdot \ctg\left(\alpha+\frac{\pi}{4}\right)\cdot \sin^{-2}\left(\alpha+\frac{\pi}{4}\right)$$
   $$=-\cos 2\alpha\cdot \frac{\cos\left(\alpha+\frac{\pi}{4}\right)}{\sin\left(\alpha+\frac{\pi}{4}\right)}\cdot \frac{1}{\sin^2\left(\alpha+\frac{\pi}{4}\right)}.$$
   Используем формулы
   $$\cos 2\alpha=\cos^2\left(\alpha+\frac{\pi}{4}\right)-\sin^2\left(\alpha+\frac{\pi}{4}\right),$$
   $$\sin\left(2\alpha+\frac{\pi}{2}\right)=2\sin\left(\alpha+\frac{\pi}{4}\right)\cos\left(\alpha+\frac{\pi}{4}\right).$$
   После упрощения получаем
   $$2.$$

Ответ

1) $$1$$;
2) $$-\frac{1}{2}\ctg^2\alpha$$;
3) $$2$$.