Упр.20.7 ГДЗ Мерзляк 10 класс Базовый уровень (Алгебра)
Упростите выражение:
- 1) $$\left(1+\mathrm{tg}\,\alpha\right)^2+\left(1-\mathrm{tg}\,\alpha\right)^2$$;
- 7) $$\frac{\mathrm{ctg}\,\alpha}{\mathrm{tg}\,\alpha+\mathrm{ctg}\,\alpha}$$;
- 2) $$\sin^4\alpha+2\sin^2\alpha\cos^2\alpha+\cos^4\alpha$$;
- 8) $$\frac{1-\mathrm{ctg}\,?}{1-\mathrm{tg}\,?}$$;
- 3) $$\frac{\sin\alpha}{1+\cos\alpha}+\frac{\sin\alpha}{1-\cos\alpha}$$;
- 9) $$\cos^4\alpha-\cos^2\alpha+\sin^2\alpha$$;
- 4) $$\mathrm{ctg}\,x+\frac{\sin x}{1+\cos\alpha}$$;
- 10) $$\sin^4\alpha+\sin^2\alpha\cos^2\alpha+\cos^2\alpha$$;
- 5) $$\frac{\sin\alpha}{1+\cos\alpha}+\frac{1+\cos\alpha}{\sin\alpha}$$;
- 11) $$\cos(-\alpha)+\cos\alpha\cdot\mathrm{tg}^2(-\alpha)$$;
- 6) $$\frac{\mathrm{tg}(3\alpha)}{\mathrm{tg}^2(3\alpha)-1}\cdot\frac{1-\mathrm{ctg}^2(3\alpha)}{\mathrm{ctg}(3\alpha)}$$;
- 12) $$\frac{1+\sin(-?)}{\cos(-?)}-\mathrm{tg}(-?)$$.
$$(1+\tg a)^2+(1-\tg a)^2=(1+2\tg a+\tg^2 a)+(1-2\tg a+\tg^2 a)=2+2\tg^2 a$$
$$2+2\tg^2 a=2(1+\tg^2 a)=\frac{2}{\cos^2 a}$$
$$\sin^4 a+2\sin^2 a\cos^2 a+\cos^4 a=(\sin^2 a+\cos^2 a)^2=1^2=1$$
$$\frac{\sin a}{1+\cos a}+\frac{\sin a}{1-\cos a} =\frac{\sin a(1-\cos a)+\sin a(1+\cos a)}{(1+\cos a)(1-\cos a)}$$
$$=\frac{\sin a(1-\cos a+1+\cos a)}{1-\cos^2 a} =\frac{2\sin a}{\sin^2 a} =\frac{2}{\sin a}$$
$$\ctg x+\frac{\sin x}{1+\cos x} =\frac{\cos x}{\sin x}+\frac{\sin x}{1+\cos x}$$
$$=\frac{\cos x(1+\cos x)+\sin^2 x}{\sin x(1+\cos x)} =\frac{\cos x+\cos^2 x+\sin^2 x}{\sin x(1+\cos x)}$$
$$=\frac{\cos x+1}{\sin x(1+\cos x)} =\frac{1}{\sin x}$$
$$\frac{\sin a}{1+\cos a}+\frac{1+\cos a}{\sin a} =\frac{\sin^2 a+(1+\cos a)^2}{(1+\cos a)\sin a}$$
$$=\frac{\sin^2 a+1+2\cos a+\cos^2 a}{(1+\cos a)\sin a} =\frac{2+2\cos a}{(1+\cos a)\sin a} =\frac{2}{\sin a}$$
$$\frac{\tg 3a}{\tg^2 3a-1}\cdot\frac{1-\ctg^2 3a}{\ctg 3a} =\frac{\tg 3a}{\tg^2 3a-1}\cdot\frac{(1-\ctg^2 3a)\tg^2 3a}{\ctg 3a\cdot\tg^2 3a}$$
$$=\frac{\tg 3a}{\tg^2 3a-1}\cdot\frac{\tg^2 3a-1}{\tg 3a}=1$$
$$\frac{\ctg a}{\tg a+\ctg a} =\frac{\ctg a\cdot\tg a}{(\tg a+\ctg a)\tg a} =\frac{1}{\tg^2 a+1}$$
$$=\frac{1}{\frac{1}{\cos^2 a}}=\cos^2 a$$
$$\frac{1-\ctg \gamma}{1-\tg \gamma} =\frac{(1-\ctg \gamma)\ctg \gamma}{(1-\tg \gamma)\ctg \gamma} =\frac{(1-\ctg \gamma)\ctg \gamma}{\ctg \gamma-1} =-\ctg \gamma$$
$$\cos^4 a-\cos^2 a+\sin^2 a =\cos^4 a-\cos^2 a+(1-\cos^2 a)$$
$$=\cos^4 a-2\cos^2 a+1 =(1-\cos^2 a)^2 =\sin^4 a$$
$$\sin^4 a+\sin^2 a\cos^2 a+\cos^2 a =\sin^4 a+\sin^2 a(1-\sin^2 a)+(1-\sin^2 a)$$
$$=\sin^4 a+\sin^2 a-\sin^4 a+1-\sin^2 a=1$$
$$\cos(-a)+\cos a\cdot\tg^2(-a)=\cos a+\cos a\cdot\tg^2 a$$
$$=\cos a(1+\tg^2 a)=\cos a\cdot\frac{1}{\cos^2 a}=\frac{1}{\cos a}$$
$$\frac{1+\sin(-\beta)}{\cos(-\beta)}-\tg(-\beta) =\frac{1-\sin\beta}{\cos\beta}+\tg\beta$$
$$=\frac{1}{\cos\beta}-\tg\beta+\tg\beta=\frac{1}{\cos\beta}$$
