Упр.469 ГДЗ Алимов 10-11 класс (Алгебра)

Упростить выражение:

1. $$\left(1+\operatorname{tg}^2 a\right)\cos^2 a-1$$
2. $$1-\sin^2 a\left(1+\operatorname{ctg} a\right)$$
3. $$1+\operatorname{tg}^2 a+\frac{1}{\sin^2 a}$$
4. $$\frac{1+\operatorname{tg}^2 a}{1+\operatorname{ctg}^2 a}$$

1) $$\left(1+\tg^2\alpha\right)\cos^2\alpha-1=\frac{\sin^2\alpha+\cos^2\alpha}{\cos^2\alpha}\cos^2\alpha-1=1-1=0.$$

Ответ: $$0$$

2) $$1-\sin^2\alpha\left(1+\ctg^2\alpha\right)=1-\sin^2\alpha\cdot\frac{\sin^2\alpha+\cos^2\alpha}{\sin^2\alpha}=1-1=0.$$

Ответ: $$0$$

3) $$1+\tg^2\alpha+\frac{1}{\sin^2\alpha}=\frac{\sin^2\alpha+\cos^2\alpha}{\cos^2\alpha}+\frac{1}{\sin^2\alpha}=\frac{1}{\cos^2\alpha}+\frac{1}{\sin^2\alpha}=\frac{\sin^2\alpha+\cos^2\alpha}{\sin^2\alpha\cos^2\alpha}=\frac{1}{\sin^2\alpha\cos^2\alpha}.$$

Ответ: $$\frac{1}{\sin^2\alpha\cos^2\alpha}$$

4) $$\frac{1+\tg^2\alpha}{1+\ctg^2\alpha}=\frac{1+\tg^2\alpha}{1+\frac{1}{\tg^2\alpha}}=\frac{1+\tg^2\alpha}{\frac{1+\tg^2\alpha}{\tg^2\alpha}}=\tg^2\alpha.$$

Ответ: $$\tg^2\alpha$$