Для нахождения значения (\cos a) воспользуемся формулой (\cos^2 a + \sin^2 a = 1):
[\cos^2 a = 1 - \sin^2 a]
[\cos^2 a = 1 - \left(\frac{3}{4}\right)^2]
[\cos^2 a = 1 - \frac{9}{16}]
[\cos^2 a = \frac{16}{16} - \frac{9}{16}]
[\cos^2 a = \frac{7}{16}]
[\cos a = \pm \sqrt{\frac{7}{16}} = \pm \frac{\sqrt{7}}{4}]

Так как по условию задачи угол (a) принадлежит третьему квадранту, то (\cos a) отрицательный:
[\cos a = -\frac{\sqrt{7}}{4}]

Теперь найдем остальные тригонометрические функции:
[\tan a = \frac{\sin a}{\cos a} = \frac{\frac{3}{4}}{-\frac{\sqrt{7}}{4}} = -\frac{3}{\sqrt{7}} = -\frac{3\sqrt{7}}{7}]

[\cot a = \frac{1}{\tan a} = -\frac{1}{-\frac{3\sqrt{7}}{7}} = \frac{7}{3\sqrt{7}} = \frac{7\sqrt{7}}{21} = \frac{\sqrt{7}}{3}]

[ \tan 2a = \frac{2\tan a}{1-\tan^2 a} = \frac{2\left(-\frac{3\sqrt{7}}{7}\right)}{1-\left(-\frac{3\sqrt{7}}{7}\right)^2} = \frac{-6\sqrt{7}}{1-\frac{3^2\cdot 7}{7^2}} = \frac{-6\sqrt{7}}{1-\frac{63}{49}} = \frac{-6\sqrt{7}}{1-\frac{63}{49}} = \frac{-6\sqrt{7}}{1-\frac{63}{49}} = \frac{-6\sqrt{7}}{\frac{49-63}{49}} = \frac{-6\sqrt{7}}{\frac{-14}{49}} = \frac{6\sqrt{7}\cdot 49}{14} = \frac{294}{14} = 21]

Итак, получаем:
[\cos a = -\frac{\sqrt{7}}{4}]
[\tan a = -\frac{3\sqrt{7}}{7}]
[\cot a = \frac{\sqrt{7}}{3}]
[\tan 2a = 21]