Упр.22.6 ГДЗ Мордкович 10-11 класс (Алгебра)
а) $$\operatorname{tg} 25+\operatorname{tg} 35$$; б) $$\operatorname{tg}\frac{\pi}{5}-\operatorname{tg}\frac{\pi}{10}$$; в) $$\operatorname{tg} 20+\operatorname{tg} 40$$; г) $$\operatorname{tg}\frac{\pi}{3}-\operatorname{tg}\frac{\pi}{4}$$.
а) Используем формулу суммы тангенсов:
$$\tg 25^\circ+\tg 35^\circ=\frac{\sin 25^\circ}{\cos 25^\circ}+\frac{\sin 35^\circ}{\cos 35^\circ} =\frac{\sin 25^\circ\cos 35^\circ+\cos 25^\circ\sin 35^\circ}{\cos 25^\circ\cos 35^\circ} =\frac{\sin(25^\circ+35^\circ)}{\cos 25^\circ\cos 35^\circ} =\frac{\sin 60^\circ}{\cos 25^\circ\cos 35^\circ} =\frac{\sqrt3}{2\cos 25^\circ\cos 35^\circ}.$$
б)
$$\tg \frac{\pi}{5}-\tg \frac{\pi}{10} =\frac{\sin \frac{\pi}{5}}{\cos \frac{\pi}{5}}-\frac{\sin \frac{\pi}{10}}{\cos \frac{\pi}{10}} =\frac{\sin \frac{\pi}{5}\cos \frac{\pi}{10}-\cos \frac{\pi}{5}\sin \frac{\pi}{10}}{\cos \frac{\pi}{5}\cos \frac{\pi}{10}} =\frac{\sin\left(\frac{\pi}{5}-\frac{\pi}{10}\right)}{\cos \frac{\pi}{5}\cos \frac{\pi}{10}} =\frac{\sin \frac{\pi}{10}}{\cos \frac{\pi}{5}\cos \frac{\pi}{10}}.$$
в)
$$\tg 20^\circ+\tg 40^\circ =\frac{\sin 20^\circ}{\cos 20^\circ}+\frac{\sin 40^\circ}{\cos 40^\circ} =\frac{\sin 20^\circ\cos 40^\circ+\cos 20^\circ\sin 40^\circ}{\cos 20^\circ\cos 40^\circ} =\frac{\sin(20^\circ+40^\circ)}{\cos 20^\circ\cos 40^\circ} =\frac{\sin 60^\circ}{\cos 20^\circ\cos 40^\circ} =\frac{\sqrt3}{2\cos 20^\circ\cos 40^\circ}.$$
г)
$$\tg \frac{\pi}{3}-\tg \frac{\pi}{4} =\frac{\sin \frac{\pi}{3}}{\cos \frac{\pi}{3}}-\frac{\sin \frac{\pi}{4}}{\cos \frac{\pi}{4}} =\frac{\sin \frac{\pi}{3}\cos \frac{\pi}{4}-\cos \frac{\pi}{3}\sin \frac{\pi}{4}}{\cos \frac{\pi}{3}\cos \frac{\pi}{4}} =\frac{\sin\left(\frac{\pi}{3}-\frac{\pi}{4}\right)}{\cos \frac{\pi}{3}\cos \frac{\pi}{4}} =\frac{\sin \frac{\pi}{12}}{\cos \frac{\pi}{3}\cos \frac{\pi}{4}} =\frac{\sin \frac{\pi}{12}}{\frac12\cdot \frac{\sqrt2}{2}} =2\sqrt2\,\sin \frac{\pi}{12}.$$
Ответ: а) $$\frac{\sqrt3}{2\cos 25^\circ\cos 35^\circ}$$; б) $$\frac{\sin \frac{\pi}{10}}{\cos \frac{\pi}{5}\cos \frac{\pi}{10}}$$; в) $$\frac{\sqrt3}{2\cos 20^\circ\cos 40^\circ}$$; г) $$2\sqrt2\,\sin \frac{\pi}{12}$$.
