Упр.23.2 ГДЗ Мерзляк 10 класс Углубленный уровень (Алгебра)

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

1. $$\begin{aligned} (1+\ctg^2\beta)+(1-\ctg\beta)^2 &=1+2\ctg\beta+\ctg^2\beta+1-2\ctg\beta+\ctg^2\beta \\ &=2+2\ctg^2\beta \\ &=2(1+\ctg^2\beta) \\ &=\frac{2}{\sin^2\beta}. \end{aligned}$$
2. $$\begin{aligned} \sin^2\alpha\cos^2\alpha(\tg^2\alpha+\ctg^2\alpha+2) &=\sin^4\alpha+\cos^4\alpha+2\sin^2\alpha\cos^2\alpha \\ &=(\sin^2\alpha+\cos^2\alpha)^2 \\ &=1. \end{aligned}$$
3. $$\begin{aligned} \frac{\cos\beta}{1-\sin\beta}+\frac{1-\sin\beta}{\cos\beta} &=\frac{\cos^2\beta+(1-\sin\beta)^2}{\cos\beta(1-\sin\beta)} \\ &=\frac{\cos^2\beta+1-2\sin\beta+\sin^2\beta}{\cos\beta(1-\sin\beta)} \\ &=\frac{2-2\sin\beta}{\cos\beta(1-\sin\beta)} \\ &=\frac{2(1-\sin\beta)}{\cos\beta(1-\sin\beta)} \\ &=\frac{2}{\cos\beta}. \end{aligned}$$
4. $$\begin{aligned} \frac{\tg^2\alpha}{1+\tg^2\alpha}\cdot\frac{1+\ctg^2\alpha}{\ctg^2\alpha} &=\frac{\tg^2\alpha}{\frac{1}{\cos^2\alpha}}\cdot\frac{\frac{1}{\sin^2\alpha}}{\ctg^2\alpha} \\ &=\tg^2\alpha. \end{aligned}$$
5. $$\begin{aligned} \cos^4\alpha+\sin^2\alpha\cos^2\alpha-\cos^2\alpha-1 &=\cos^2\alpha(\cos^2\alpha+\sin^2\alpha)-\cos^2\alpha-1 \\ &=\cos^2\alpha-\cos^2\alpha-1 \\ &=-1. \end{aligned}$$
6. $$\begin{aligned} \tg(-\alpha)\ctg\alpha+\sin^2(-\alpha) &=-\tg\alpha\ctg\alpha+\sin^2\alpha \\ &=-1+\sin^2\alpha \\ &=-\cos^2\alpha. \end{aligned}$$

Ответ

1) $$\frac{2}{\sin^2\beta}$$; 2) $$1$$; 3) $$\frac{2}{\cos\beta}$$; 4) $$\tg^2\alpha$$; 5) $$-1$$; 6) $$-\cos^2\alpha$$.