Упр.25.9 ГДЗ Мерзляк 10 класс Углубленный уровень (Алгебра)

Упростите выражение:

1. 1) $$\cos^2\left(\frac{\pi}{4}+\alpha\right)+\cos^2\left(\frac{\pi}{4}-\alpha\right)+\sin\left(\frac{3\pi}{2}-\alpha\right)\cos\left(\frac{3\pi}{2}+\alpha\right)\operatorname{tg}(\pi+\alpha);$$
   2) $$\frac{\cos^2(20^\circ-\alpha)}{\sin^2(70^\circ+\alpha)}+\operatorname{tg}(\alpha+10^\circ)\operatorname{ctg}(80^\circ-\alpha).$$

$$\cos^2\left(\frac{\pi}{4}+\alpha\right)+\cos^2\left(\frac{\pi}{4}-\alpha\right)+\sin\left(\frac{3\pi}{2}-\alpha\right)\cos\left(\frac{3\pi}{2}+\alpha\right)\tg(\pi+\alpha)$$

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

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

$$=1-\sin^2\alpha=\cos^2\alpha.$$

$$\frac{\cos^2(20^\circ-\alpha)}{\sin^2(70^\circ+\alpha)}+\tg(\alpha+10^\circ)\cdot\ctg(80^\circ-\alpha)$$

$$=\frac{\cos^2\bigl(90^\circ-(70^\circ+\alpha)\bigr)}{\sin^2(70^\circ+\alpha)}+\tg(\alpha+10^\circ)\cdot\ctg\bigl(90^\circ-(10^\circ+\alpha)\bigr)$$

$$=\frac{\sin^2(70^\circ+\alpha)}{\sin^2(70^\circ+\alpha)}+\tg^2(10^\circ+\alpha)$$

$$=1+\tg^2(\alpha+10^\circ)=\frac{1}{\cos^2(\alpha+10^\circ)}.$$

Ответ: 1) $$\cos^2\alpha$$; 2) $$\dfrac{1}{\cos^2(\alpha+10^\circ)}$$.