Дано: $$\sin a = \frac{3}{5}$$, $$0 < a < 90^{\circ}$$
Найдем $$\cos a$$:
Используем основное тригонометрическое тождество:$$\sin^2 a + \cos^2 a = 1$$$$\cos^2 a = 1 - \sin^2 a = 1 - (\frac{3}{5})^2 = 1 - \frac{9}{25} = \frac{25 - 9}{25} = \frac{16}{25}$$Так как $$0 < a < 90^{\circ}$$, то $$\cos a > 0$$, поэтому:$$\cos a = \sqrt{\frac{16}{25}} = \frac{4}{5}$$
Найдем $$\tan a$$:
$$\tan a = \frac{\sin a}{\cos a} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{5} \cdot \frac{5}{4} = \frac{3}{4}$$
Найдем $$\cot a$$:
$$\cot a = \frac{1}{\tan a} = \frac{1}{\frac{3}{4}} = \frac{4}{3}$$