Упр.20.8 ГДЗ Мерзляк 10 класс Базовый уровень (Алгебра)
1) (1+ctg ?)^2+(1-ctg ?)^2; 5) tg^2(a)/(1+tg^2(a))·(1+ctg^2(a))/ctg^2(a);
2) sin^2(a)cos^2(a)(tg^2(a)+ctg^2(a)+2); 6) (1+tg a)/(1+ctg a);
3) tg x+(cos x)/(1+sin x); 7) cos^4(a)+sin^2(a)cos^2(a)-cos^2(a)-1;
4) (cos ?)/(1-sin ?)+(1-sin ?)/(cos ?); 8) tg(-a)ctg(a)+sin^2(-a).
$$\left(1+\ctg \beta\right)^2+\left(1-\ctg \beta\right)^2$$
$$=1+2\ctg \beta+\ctg^2 \beta+1-2\ctg \beta+\ctg^2 \beta$$
$$=2+2\ctg^2 \beta=2\left(1+\ctg^2 \beta\right)$$
$$=2\cdot \frac{1}{\sin^2 \beta}=\frac{2}{\sin^2 \beta}.$$$$\sin^2 a\cos^2 a\left(\tg^2 a+\ctg^2 a+2\right)$$
$$=\sin^2 a\cos^2 a\left(\frac{1}{\cos^2 a}+\frac{1}{\sin^2 a}\right)$$
$$=\sin^2 a+\cos^2 a=1.$$$$\tg x+\frac{\cos x}{1+\sin x}=\frac{\sin x}{\cos x}+\frac{\cos x}{1+\sin x}$$
$$=\frac{\sin x(1+\sin x)+\cos^2 x}{\cos x(1+\sin x)}$$
$$=\frac{\sin x+\sin^2 x+\cos^2 x}{\cos x(1+\sin x)}$$
$$=\frac{\sin x+1}{\cos x(1+\sin x)}=\frac{1}{\cos x}.$$$$\frac{\cos \beta}{1-\sin \beta}+\frac{1-\sin \beta}{\cos \beta}$$
$$=\frac{\cos^2 \beta+(1-\sin \beta)^2}{(1-\sin \beta)\cos \beta}$$
$$=\frac{\cos^2 \beta+1-2\sin \beta+\sin^2 \beta}{(1-\sin \beta)\cos \beta}$$
$$=\frac{2-2\sin \beta}{(1-\sin \beta)\cos \beta}=\frac{2}{\cos \beta}.$$$$\frac{\tg^2 a}{1+\tg^2 a}\cdot \frac{1+\ctg^2 a}{\ctg^2 a}$$
$$=\frac{\tg^2 a}{1+\tg^2 a}\cdot \frac{(1+\ctg^2 a)\tg^2 a}{\ctg^2 a\cdot \tg^2 a}$$
$$=\frac{\tg^2 a}{1+\tg^2 a}\cdot \frac{\tg^2 a+1}{1}=\tg^2 a.$$$$\frac{1+\tg a}{1+\ctg a}=\frac{(1+\tg a)\tg a}{(1+\ctg a)\tg a}$$
$$=\frac{(1+\tg a)\tg a}{\tg a+1}=\tg a.$$$$\cos^4 a+\sin^2 a\cos^2 a-\cos^2 a-1$$
$$=\cos^4 a+(1-\cos^2 a)\cos^2 a-\cos^2 a-1$$
$$=\cos^4 a+\cos^2 a-\cos^4 a-\cos^2 a-1=-1.$$$$\tg(-a)\ctg a+\sin^2(-a)=-\tg a\cdot \ctg a+\sin^2 a$$
$$=-1+\sin^2 a=\sin^2 a-1=-\cos^2 a.$$
Ответ
1) $$\frac{2}{\sin^2 \beta}$$; 2) $$1$$; 3) $$\frac{1}{\cos x}$$; 4) $$\frac{2}{\cos \beta}$$; 5) $$\tg^2 a$$; 6) $$\tg a$$; 7) $$-1$$; 8) $$-\cos^2 a$$.
