Вычислите значения тригонометрических выражений с тангенсом и котангенсом.

Для решения этих заданий воспользуемся значениями тригонометрических функций из таблицы и формулами приведения. **92.** Вычислите: а) $\text{tg}\frac{5\pi}{4} = \text{tg}(\pi + \frac{\pi}{4}) = \text{tg}\frac{\pi}{4} = \mathbf{1}$ б) $\text{ctg}\frac{4\pi}{3} = \text{ctg}(\pi + \frac{\pi}{3}) = \text{ctg}\frac{\pi}{3} = \mathbf{\frac{\sqrt{3}}{3}}$ в) $\text{tg}\frac{5\pi}{6} = \text{tg}(\pi - \frac{\pi}{6}) = -\text{tg}\frac{\pi}{6} = \mathbf{-\frac{\sqrt{3}}{3}}$ г) $\text{ctg}\frac{11\pi}{6} = \text{ctg}(2\pi - \frac{\pi}{6}) = -\text{ctg}\frac{\pi}{6} = \mathbf{-\sqrt{3}}$ **93.** а) $\text{tg}(-\frac{5\pi}{4}) = -\text{tg}\frac{5\pi}{4} = -\text{tg}(\pi + \frac{\pi}{4}) = -\text{tg}\frac{\pi}{4} = \mathbf{-1}$ б) $\text{ctg}(-\frac{\pi}{3}) = -\text{ctg}\frac{\pi}{3} = \mathbf{-\frac{\sqrt{3}}{3}}$ в) $\text{tg}(-\frac{\pi}{6}) = \mathbf{-\frac{\sqrt{3}}{3}}$ г) $\text{ctg}(-\frac{2\pi}{3}) = -\text{ctg}\frac{2\pi}{3} = -\text{ctg}(\pi - \frac{\pi}{3}) = -(-\text{ctg}\frac{\pi}{3}) = \text{ctg}\frac{\pi}{3} = \mathbf{\frac{\sqrt{3}}{3}}$ **94.** а) $\text{tg}\frac{\pi}{4} + \text{ctg}\frac{5\pi}{4} = 1 + \text{ctg}(\pi + \frac{\pi}{4}) = 1 + 1 = \mathbf{2}$ б) $\text{ctg}\frac{\pi}{3} \cdot \text{tg}\frac{\pi}{6} = \frac{\sqrt{3}}{3} \cdot \frac{\sqrt{3}}{3} = \frac{3}{9} = \mathbf{\frac{1}{3}}$ в) $\text{tg}\frac{\pi}{6} - \text{ctg}\frac{\pi}{6} = \frac{\sqrt{3}}{3} - \sqrt{3} = \frac{\sqrt{3} - 3\sqrt{3}}{3} = \mathbf{-\frac{2\sqrt{3}}{3}}$ г) $\text{tg}\frac{9\pi}{4} + \text{ctg}\frac{\pi}{4} = \text{tg}(2\pi + \frac{\pi}{4}) + 1 = 1 + 1 = \mathbf{2}$ **95.** а) $\text{tg}\frac{\pi}{4} \cdot \sin\frac{\pi}{3} \cdot \text{ctg}\frac{\pi}{6} = 1 \cdot \frac{\sqrt{3}}{2} \cdot \sqrt{3} = \frac{3}{2} = \mathbf{1,5}$ б) $2\sin\pi + 3\cos\pi + \text{ctg}\frac{\pi}{2} = 2 \cdot 0 + 3 \cdot (-1) + 0 = \mathbf{-3}$ в) $2\sin\frac{\pi}{3} \cdot \cos\frac{\pi}{6} - \frac{1}{2}\text{tg}\frac{\pi}{3} = 2 \cdot \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{1}{2} \cdot \sqrt{3} = \frac{3}{2} - \frac{\sqrt{3}}{2} = \mathbf{\frac{3 - \sqrt{3}}{2}}$ г) $2\text{tg}0 + 8\cos\frac{3\pi}{2} - 6\sin\frac{\pi}{3} = 2 \cdot 0 + 8 \cdot 0 - 6 \cdot \frac{\sqrt{3}}{2} = \mathbf{-3\sqrt{3}}$ **96.** Воспользуемся свойством $\text{tg}\alpha \cdot \text{ctg}\alpha = 1$: а) $\text{tg}\frac{\pi}{5} \cdot \text{ctg}\frac{\pi}{5} = \mathbf{1}$ б) $3\text{tg}2,3 \cdot \text{ctg}2,3 = 3 \cdot 1 = \mathbf{3}$ в) $\text{tg}\frac{\pi}{7} \cdot \text{ctg}\frac{\pi}{7} = \mathbf{1}$ г) $7\text{tg}\frac{\pi}{12} \cdot \text{ctg}\frac{\pi}{12} = 7 \cdot 1 = \mathbf{7}$