Упр.22.13 ГДЗ Мерзляк 10 класс Базовый уровень (Алгебра)

1)

$$\left(\tg \frac{5\pi}{4}+\tg\left(\frac{\pi}{2}-a\right)\right)^2+\left(\ctg \frac{9\pi}{4}+\ctg(\pi-a)\right)^2$$
$$=\left(\tg \left(\pi+\frac{\pi}{4}\right)+\ctg a\right)^2+\left(\ctg \left(2\pi+\frac{\pi}{4}\right)-\ctg a\right)^2$$
$$=\left(\tg \frac{\pi}{4}+\ctg a\right)^2+\left(\ctg \frac{\pi}{4}-\ctg a\right)^2$$
$$=(1+\ctg a)^2+(1-\ctg a)^2$$
$$=1+2\ctg a+\ctg^2 a+1-2\ctg a+\ctg^2 a$$
$$=2\left(1+\ctg^2 a\right)=\frac{2}{\sin^2 a}.$$

Тождество доказано.

2)

$$\frac{\cos^2\left(\frac{\pi}{3}+a\right)}{\tg^2\left(\frac{\pi}{6}-a\right)}+\sin^2\left(\frac{\pi}{3}+a\right)\tg^2\left(\frac{\pi}{6}-a\right)$$
$$=\cos^2\left(\frac{\pi}{2}-\left(\frac{\pi}{6}-a\right)\right)\cdot \frac{\sin^2\left(\frac{\pi}{6}-a\right)}{\cos^2\left(\frac{\pi}{6}-a\right)} +\sin^2\left(\frac{\pi}{2}-\left(\frac{\pi}{6}-a\right)\right)\cdot \frac{\sin^2\left(\frac{\pi}{6}-a\right)}{\cos^2\left(\frac{\pi}{6}-a\right)}$$
$$=\sin^2\left(\frac{\pi}{6}-a\right)+\cos^2\left(\frac{\pi}{6}-a\right) =1.$$

Тождество доказано.

3)

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

Тождество доказано.