Упр.1298 ГДЗ Алимов 10-11 класс (Алгебра)

$$\frac{\tg \alpha+\tg \beta}{\ctg \alpha+\ctg \beta}=\frac{\tg \alpha+\tg \beta}{\frac{1}{\tg \alpha}+\frac{1}{\tg \beta}}$$

$$=\frac{\tg \alpha+\tg \beta}{\frac{\tg \alpha+\tg \beta}{\tg \alpha \cdot \tg \beta}}=\tg \alpha \cdot \tg \beta.$$

Ответ: $$\tg \alpha \cdot \tg \beta.$$

$$(\sin \alpha+\cos \alpha)^2+(\sin \alpha-\cos \alpha)^2$$

$$=\sin^2 \alpha+2\sin \alpha \cos \alpha+\cos^2 \alpha+\sin^2 \alpha-2\sin \alpha \cos \alpha+\cos^2 \alpha$$

$$=2\sin^2 \alpha+2\cos^2 \alpha=2(\sin^2 \alpha+\cos^2 \alpha)=2.$$

Ответ: $$2.$$

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

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

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

$$=\frac{\sqrt{2}\cos \alpha}{\sqrt{2}\sin \alpha}=\ctg \alpha.$$

Ответ: $$\ctg \alpha.$$

$$\frac{\sin \alpha+2\sin\left(\frac{\pi}{3}-\alpha\right)}{2\cos\left(\frac{\pi}{6}-\alpha\right)-\sqrt{3}\cos \alpha}$$

$$\sin\left(\frac{\pi}{3}-\alpha\right)=\sin\frac{\pi}{3}\cos \alpha-\cos\frac{\pi}{3}\sin \alpha=\frac{\sqrt{3}}{2}\cos \alpha-\frac{1}{2}\sin \alpha,$$

$$\cos\left(\frac{\pi}{6}-\alpha\right)=\cos\frac{\pi}{6}\cos \alpha+\sin\frac{\pi}{6}\sin \alpha=\frac{\sqrt{3}}{2}\cos \alpha+\frac{1}{2}\sin \alpha.$$

Подставим:
$$\frac{\sin \alpha+2\left(\frac{\sqrt{3}}{2}\cos \alpha-\frac{1}{2}\sin \alpha\right)}{2\left(\frac{\sqrt{3}}{2}\cos \alpha+\frac{1}{2}\sin \alpha\right)-\sqrt{3}\cos \alpha}$$

$$=\frac{\sqrt{3}\cos \alpha}{\sin \alpha}=\sqrt{3}\ctg \alpha.$$

Ответ: $$\sqrt{3}\ctg \alpha.$$