Упр.8 ГДЗ Колмогоров 10-11 класс (Алгебра)
а) $$\cos^2\alpha-\cos^4\alpha+\sin^4\alpha$$
$$\cos^2\alpha-\cos^4\alpha+\sin^4\alpha = \cos^2\alpha+(\sin^2\alpha)^2-(\cos^2\alpha)^2$$
$$= \cos^2\alpha+(\sin^2\alpha+\cos^2\alpha)(\sin^2\alpha-\cos^2\alpha)$$
$$= \cos^2\alpha+\sin^2\alpha-\cos^2\alpha = \sin^2\alpha$$
б) $$\dfrac{1-2\cos^2\beta}{\cos\beta+\sin\beta}$$
$$\dfrac{1-2\cos^2\beta}{\cos\beta+\sin\beta} = \dfrac{\sin^2\beta+\cos^2\beta-2\cos^2\beta}{\cos\beta+\sin\beta} = \dfrac{\sin^2\beta-\cos^2\beta}{\cos\beta+\sin\beta}$$
$$= \dfrac{(\sin\beta-\cos\beta)(\sin\beta+\cos\beta)}{\cos\beta+\sin\beta} = \sin\beta-\cos\beta$$
При $$\cos\beta+\sin\beta\neq 0$$, то есть $$\beta\neq -\dfrac{\pi}{4}+\pi n,\ n\in\mathbb Z.$$
в) $$\left(\sin^2\alpha+\tg^2\alpha\sin^2\alpha\right)\ctg\alpha$$
$$\left(\sin^2\alpha+\tg^2\alpha\sin^2\alpha\right)\ctg\alpha = \sin^2\alpha\left(1+\tg^2\alpha\right)\ctg\alpha$$
$$= \sin^2\alpha\cdot \dfrac{1}{\cos^2\alpha}\cdot \dfrac{\cos\alpha}{\sin\alpha} = \tg\alpha$$
г) $$\dfrac{\sin^2 t-1}{\cos^4 t}+\tg^2 t$$
$$\dfrac{\sin^2 t-1}{\cos^4 t}+\tg^2 t = \dfrac{\sin^2 t-(\sin^2 t+\cos^2 t)}{\cos^4 t}+\dfrac{\sin^2 t}{\cos^2 t}$$
$$= \dfrac{-\cos^2 t}{\cos^4 t}+\dfrac{\sin^2 t}{\cos^2 t} = -\dfrac{1}{\cos^2 t}+\dfrac{\sin^2 t}{\cos^2 t}$$
$$= \dfrac{-(\sin^2 t+\cos^2 t)+\sin^2 t}{\cos^2 t} = -\dfrac{\cos^2 t}{\cos^2 t} = -1$$
Ответ: а) $$\sin^2\alpha$$; б) $$\sin\beta-\cos\beta$$; в) $$\tg\alpha$$; г) $$-1$$.
