Преобразуй выражение ctg a sin (-α) cos (-a)

- a) $ctg \alpha \cdot sin(-\alpha) - cos(-\alpha) = ctg \alpha \cdot (-sin \alpha) - cos \alpha = -ctg \alpha \cdot sin \alpha - cos \alpha = -\frac{cos \alpha}{sin \alpha} \cdot sin \alpha - cos \alpha = -cos \alpha - cos \alpha = -2cos \alpha$ - б) $\frac{1 - sin^2(-x)}{cos x} = \frac{1 - (-sin x)^2}{cos x} = \frac{1 - sin^2 x}{cos x} = \frac{cos^2 x}{cos x} = cos x$ - в) $tg(-\beta) \cdot ctg \beta + sin^2 \beta = -tg \beta \cdot ctg \beta + sin^2 \beta = -tg \beta \cdot \frac{1}{tg \beta} + sin^2 \beta = -1 + sin^2 \beta = -cos^2 \beta$ - г) $\frac{tg(-x) + 1}{1 - ctg x} = \frac{-tg(x) + 1}{1 - ctg x} = \frac{-\frac{sin x}{cos x} + 1}{1 - \frac{cos x}{sin x}} = \frac{\frac{-sin x + cos x}{cos x}}{\frac{sin x - cos x}{sin x}} = \frac{(cos x - sin x) \cdot sin x}{cos x \cdot (sin x - cos x)} = -\frac{sin x}{cos x} = -tg x$ *Перевод:* - *а) ctg α sin (-α) - cos (-α) = ctg α (-sin α) - cos α = -ctg α sin α - cos α = -cos α/sin α sin α - cos α = -cos α - cos α = -2cos α* - *б) (1 - sin²(-x)) / cos x = (1 - (-sin x)²) / cos x = (1 - sin² x) / cos x = cos² x / cos x = cos x* - *в) tg(-β) ctg β + sin² β = -tg β ctg β + sin² β = -tg β (1/tg β) + sin² β = -1 + sin² β = -cos² β* - *г) (tg(-x) + 1) / (1 - ctg x) = (-tg(x) + 1) / (1 - ctg x) = (-sin x/cos x + 1) / (1 - cos x/sin x) = ((-sin x + cos x) / cos x) / ((sin x - cos x) / sin x) = ((cos x - sin x) sin x) / (cos x (sin x - cos x)) = -sin x / cos x = -tg x*