20. Вычислите √3(ctg 13π/12 - ctg 5π/12)

Преобразуем котангенсы, используя формулы приведения:

\(\ctg(\frac{13\pi}{12}) = \ctg(\pi + \frac{\pi}{12}) = \ctg(\frac{\pi}{12})\) \(\ctg(\frac{5\pi}{12}) = \ctg(\frac{\pi}{2} - \frac{\pi}{12}) = \tan(\frac{\pi}{12})\) Тогда выражение примет вид: \(\sqrt{3}(\ctg(\frac{\pi}{12}) - \tan(\frac{\pi}{12})) = \sqrt{3}(\frac{\cos(\frac{\pi}{12})}{\sin(\frac{\pi}{12})} - \frac{\sin(\frac{\pi}{12})}{\cos(\frac{\pi}{12})}) = \sqrt{3}(\frac{\cos^2(\frac{\pi}{12}) - \sin^2(\frac{\pi}{12})}{\sin(\frac{\pi}{12})\cos(\frac{\pi}{12})})\) Используем формулы двойного угла: \(\cos^2 x - \sin^2 x = \cos 2x\) \(2 \sin x \cos x = \sin 2x\) Тогда: \(\sqrt{3}(\frac{\cos(\frac{\pi}{6})}{\frac{1}{2}\sin(\frac{\pi}{6})}) = 2\sqrt{3}\frac{\cos(\frac{\pi}{6})}{\sin(\frac{\pi}{6})} = 2\sqrt{3}\ctg(\frac{\pi}{6}) = 2\sqrt{3} \cdot \sqrt{3} = 2 \cdot 3 = 6\)

Ответ: 6