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

Рассмотрим вариант решения задания из учебника Алимов, Колягин, Ткачёва 11 класс, Просвещение: 1298

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

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

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

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

Ответ

1) $$\tg \alpha \cdot \tg \beta$$
2) $$2$$
3) $$\tg \alpha$$
4) $$\sqrt{3}\ctg \alpha$$