Упр.23.2 ГДЗ Мерзляк 10 класс Углубленный уровень (Алгебра)
Упростите выражение:
- $$\left(1+\operatorname{ctg}(?)\right)^2+\left(1-\operatorname{ctg}(?)\right)^2$$;
- $$\sin^2(?)\cos^2(?)\left(\operatorname{tg}^2(?) + \operatorname{ctg}^2(?) + 2\right)$$;
- $$\frac{\cos(?)}{1-\sin(?)}+\frac{1-\sin(?)}{\cos(?)}$$;
- $$\frac{\operatorname{tg}^2(?)}{1+\operatorname{tg}^2(?)}\cdot\frac{1+\operatorname{ctg}^2(?)}{\operatorname{ctg}^2(?)}$$;
- $$\cos^4(?) + \sin^2(?)\cos^2(?) — \cos^2(?) — 1$$;
- $$\operatorname{tg}(-?)\operatorname{ctg}(?) + \sin^2(-?)$$.
$$(1+\ctg \beta)^2+(1-\ctg \beta)^2$$
$$=1+2\ctg \beta+\ctg^2 \beta+1-2\ctg \beta+\ctg^2 \beta$$
$$=2+2\ctg^2 \beta=2(1+\ctg^2 \beta)=\frac{2}{\sin^2 \beta}$$
$$\sin^2 \alpha \cos^2 \alpha \, (\tg^2 \alpha+\ctg^2 \alpha+2)$$
$$=\sin^4 \alpha+\cos^4 \alpha+2\sin^2 \alpha \cos^2 \alpha$$
$$=(\sin^2 \alpha+\cos^2 \alpha)^2=1$$
$$\frac{\cos \beta}{1-\sin \beta}+\frac{1-\sin \beta}{\cos \beta}$$
$$=\frac{\cos^2 \beta+(1-\sin \beta)^2}{\cos \beta(1-\sin \beta)}$$
$$=\frac{\cos^2 \beta+1-2\sin \beta+\sin^2 \beta}{\cos \beta(1-\sin \beta)}$$
$$=\frac{2-2\sin \beta}{\cos \beta(1-\sin \beta)} =\frac{2(1-\sin \beta)}{\cos \beta(1-\sin \beta)} =\frac{2}{\cos \beta}$$
$$\frac{\tg^2 \alpha}{1+\tg^2 \alpha}\cdot \frac{1+\ctg^2 \alpha}{\ctg^2 \alpha}$$
$$=\frac{\tg^2 \alpha}{\frac{1}{\cos^2 \alpha}}\cdot \frac{\frac{1}{\sin^2 \alpha}}{\ctg^2 \alpha} =\frac{\tg^2 \alpha \cos^2 \alpha}{\sin^2 \alpha \ctg^2 \alpha}$$
$$=\frac{\frac{\sin^2 \alpha}{\cos^2 \alpha}\cos^2 \alpha}{\sin^2 \alpha \cdot \frac{\cos^2 \alpha}{\sin^2 \alpha}} =\tg^2 \alpha$$
$$\cos^4 \alpha+\sin^2 \alpha \cos^2 \alpha-\cos^2 \alpha-1$$
$$=\cos^2 \alpha(\cos^2 \alpha+\sin^2 \alpha)-\cos^2 \alpha-1$$
$$=\cos^2 \alpha-\cos^2 \alpha-1=-1$$
$$\tg(-\alpha)\ctg \alpha+\sin^2(-\alpha)$$
$$=-\tg \alpha \cdot \ctg \alpha+\sin^2 \alpha =-1+\sin^2 \alpha$$
$$=-(1-\sin^2 \alpha)=-\cos^2 \alpha$$
