Упр.52 Повторение ГДЗ Колмогоров 10-11 класс (Алгебра)

Задача

а) tg^2(?)-sin^2(?)-tg^2(?)sin^2(?);
б) v(sin^2(?)(1+ctg(?))+cos^2(?)(1+tg(?)));
в) (3sin(?)+2cos(?))^2+(2sin(?)-3cos(?))^2;
г) cos(?)tg(?)/sin^2(?)-ctg(?)cos(?).

Подробный ответ

а)

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

б)

$$ \sqrt{\sin^2\beta(1+\ctg\beta)+\cos^2\beta(1+\tg\beta)} $$
$$ =\sqrt{\sin^2\beta+\sin^2\beta\ctg\beta+\cos^2\beta+\cos^2\beta\tg\beta} $$
$$ =\sqrt{\sin^2\beta+\sin\beta\cos\beta+\cos^2\beta+\sin\beta\cos\beta} $$
$$ =\sqrt{(\sin\beta+\cos\beta)^2} =|\sin\beta+\cos\beta|. $$

в)

$$ (3\sin\alpha+2\cos\alpha)^2+(2\sin\alpha-3\cos\alpha)^2 $$
$$ =9\sin^2\alpha+12\sin\alpha\cos\alpha+4\cos^2\alpha +4\sin^2\alpha-12\sin\alpha\cos\alpha+9\cos^2\alpha $$
$$ =13\sin^2\alpha+13\cos^2\alpha =13(\sin^2\alpha+\cos^2\alpha)=13. $$

г)

$$ \frac{\cos\beta\cdot\tg\beta}{\sin^2\beta}-\ctg\beta\cdot\cos\beta $$
$$ =\frac{\cos\beta\cdot\frac{\sin\beta}{\cos\beta}}{\sin^2\beta}-\frac{\cos\beta}{\sin\beta}\cdot\cos\beta $$
$$ =\frac{\sin\beta}{\sin^2\beta}-\frac{\cos^2\beta}{\sin\beta} =\frac{1-\cos^2\beta}{\sin\beta} =\frac{\sin^2\beta}{\sin\beta} =\sin\beta. $$