Упр.771 ГДЗ Колягин Ткачёва 11 класс (Алгебра)

2) 2sina + 2sin(пи/3 — a)/3cos(пи/6 — a) — корень 3 cosa;
3) tga + tg бета /ctga + ctg бета;
4) (sina + cosa_2 + (sina — cosa)2.

1. $$\frac{\sin\left(\frac{\pi}{4}+a\right)-\cos\left(\frac{\pi}{4}+a\right)}{\sin\left(\frac{\pi}{4}+a\right)+\cos\left(\frac{\pi}{4}+a\right)}$$
   $$=\frac{\sin\frac{\pi}{4}\cos a+\cos\frac{\pi}{4}\sin a-\left(\cos\frac{\pi}{4}\cos a-\sin\frac{\pi}{4}\sin a\right)}{\sin\frac{\pi}{4}\cos a+\cos\frac{\pi}{4}\sin a+\cos\frac{\pi}{4}\cos a-\sin\frac{\pi}{4}\sin a}$$
   $$=\frac{\frac{\sqrt2}{2}\cos a+\frac{\sqrt2}{2}\sin a-\frac{\sqrt2}{2}\cos a+\frac{\sqrt2}{2}\sin a}{\frac{\sqrt2}{2}\cos a+\frac{\sqrt2}{2}\sin a+\frac{\sqrt2}{2}\cos a-\frac{\sqrt2}{2}\sin a}$$
   $$=\frac{\sqrt2\sin a}{\sqrt2\cos a}=\tg a.$$

2. $$\frac{\sin a+2\sin\left(\frac{\pi}{3}-a\right)}{2\cos\left(\frac{\pi}{6}-a\right)-\sqrt3\cos a}$$
   $$=\frac{\sin a+2\left(\sin\frac{\pi}{3}\cos a-\cos\frac{\pi}{3}\sin a\right)}{2\left(\cos\frac{\pi}{6}\cos a+\sin\frac{\pi}{6}\sin a\right)-\sqrt3\cos a}$$
   $$=\frac{\sin a+2\left(\frac{\sqrt3}{2}\cos a-\frac12\sin a\right)}{2\left(\frac{\sqrt3}{2}\cos a+\frac12\sin a\right)-\sqrt3\cos a}$$
   $$=\frac{\sqrt3\cos a}{\sin a}=\sqrt3\,\ctg a.$$

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

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

Ответ

1) $$\tg a$$; 2) $$\sqrt3\,\ctg a$$; 3) $$\tg a\cdot \tg \beta$$; 4) $$2$$.