Упр.1299 ГДЗ Алимов 10-11 класс (Алгебра)

1)

$$\frac{1-\tg^2\left(\frac{\pi}{4}-\alpha\right)}{1+\tg^2\left(\frac{\pi}{4}-\alpha\right)} = \frac{1-\dfrac{\sin^2\left(\frac{\pi}{4}-\alpha\right)}{\cos^2\left(\frac{\pi}{4}-\alpha\right)}}{1+\dfrac{\sin^2\left(\frac{\pi}{4}-\alpha\right)}{\cos^2\left(\frac{\pi}{4}-\alpha\right)}} = \frac{\cos^2\left(\frac{\pi}{4}-\alpha\right)-\sin^2\left(\frac{\pi}{4}-\alpha\right)}{\cos^2\left(\frac{\pi}{4}-\alpha\right)+\sin^2\left(\frac{\pi}{4}-\alpha\right)}$$
$$= \cos\left(2\left(\frac{\pi}{4}-\alpha\right)\right) = \cos\left(\frac{\pi}{2}-2\alpha\right) = \sin 2\alpha$$

Ответ: $$\sin 2\alpha$$.

2)

$$\frac{\sin 2\alpha}{1+\cos 2\alpha} = \frac{2\sin\alpha\cos\alpha}{\cos^2\alpha+\sin^2\alpha+\cos^2\alpha-\sin^2\alpha} = \frac{2\sin\alpha\cos\alpha}{2\cos^2\alpha} = \frac{\sin\alpha}{\cos\alpha} = \tg\alpha$$

Ответ: $$\tg\alpha$$.