Упр.22.6 ГДЗ Мордкович 10-11 класс (Алгебра)
a) tg 25 + tg 35;
6) tg РїРё/5 — tg РїРё/10;
РІ) tg 20 + tg 40;
Рі) tg РїРё/3 — tg РїРё/4.
Решение
$$\tg 25^\circ+\tg 35^\circ=\frac{\sin 25^\circ}{\cos 25^\circ}+\frac{\sin 35^\circ}{\cos 35^\circ}$$
$$=\frac{\sin 25^\circ\cos 35^\circ+\cos 25^\circ\sin 35^\circ}{\cos 25^\circ\cos 35^\circ}=\frac{\sin(25^\circ+35^\circ)}{\cos 25^\circ\cos 35^\circ}$$
$$=\frac{\sin 60^\circ}{\cos 25^\circ\cos 35^\circ}=\frac{\sqrt{3}}{2\cos 25^\circ\cos 35^\circ}$$
$$\tg \frac{\pi}{5}-\tg \frac{\pi}{10}=\frac{\sin \frac{\pi}{5}}{\cos \frac{\pi}{5}}-\frac{\sin \frac{\pi}{10}}{\cos \frac{\pi}{10}}$$
$$=\frac{\sin \frac{\pi}{5}\cos \frac{\pi}{10}-\cos \frac{\pi}{5}\sin \frac{\pi}{10}}{\cos \frac{\pi}{5}\cos \frac{\pi}{10}}=\frac{\sin\left(\frac{\pi}{5}-\frac{\pi}{10}\right)}{\cos \frac{\pi}{5}\cos \frac{\pi}{10}}$$
$$=\frac{\sin \frac{\pi}{10}}{\cos \frac{\pi}{5}\cos \frac{\pi}{10}}$$
$$\tg 20^\circ+\tg 40^\circ=\frac{\sin 20^\circ}{\cos 20^\circ}+\frac{\sin 40^\circ}{\cos 40^\circ}$$
$$=\frac{\sin 20^\circ\cos 40^\circ+\cos 20^\circ\sin 40^\circ}{\cos 20^\circ\cos 40^\circ}=\frac{\sin(20^\circ+40^\circ)}{\cos 20^\circ\cos 40^\circ}$$
$$=\frac{\sin 60^\circ}{\cos 20^\circ\cos 40^\circ}=\frac{\sqrt{3}}{2\cos 20^\circ\cos 40^\circ}$$
$$\tg \frac{\pi}{3}-\tg \frac{\pi}{4}=\frac{\sin \frac{\pi}{3}}{\cos \frac{\pi}{3}}-\frac{\sin \frac{\pi}{4}}{\cos \frac{\pi}{4}}$$
$$=\frac{\sin \frac{\pi}{3}\cos \frac{\pi}{4}-\cos \frac{\pi}{3}\sin \frac{\pi}{4}}{\cos \frac{\pi}{3}\cos \frac{\pi}{4}}=\frac{\sin\left(\frac{\pi}{3}-\frac{\pi}{4}\right)}{\cos \frac{\pi}{3}\cos \frac{\pi}{4}}$$
$$=\frac{\sin \frac{\pi}{12}}{\cos \frac{\pi}{3}\cos \frac{\pi}{4}}=\frac{\sin \frac{\pi}{12}}{\frac12\cdot \frac{\sqrt2}{2}}=2\sqrt2\,\sin \frac{\pi}{12}$$
Ответ
а) $$\frac{\sqrt{3}}{2\cos 25^\circ\cos 35^\circ}$$;
б) $$\frac{\sin \frac{\pi}{10}}{\cos \frac{\pi}{5}\cos \frac{\pi}{10}}$$;
в) $$\frac{\sqrt{3}}{2\cos 20^\circ\cos 40^\circ}$$;
г) $$2\sqrt2\,\sin \frac{\pi}{12}$$.
