Найдите tg a и ctg a, если sin a = 3/5

Из формулы sin^2a + cos ^2a = 1, выразим

cos ^2a = 1 - sin^2a;

cos ^2a = 1 - (3/5) ^2;

cos ^2a = 1 - 9/25;

cos ^2a = 25/25 - 9/25;

cos ^2a = 16/25;

cos a = 4/5;

tg a = sin a/cos a;

tg a = 3/5 : 4/5;

tg a = 3/5 * 5/4;

tg a = (3 * 5) / (5 * 4);

tg a = (3 * 1) / (1 * 4);

tg a = 3/4;

ctg a = cos a/sin a;

сtg a = 4/5 : 3/5;

сtg a = 4/5 * 5/3;

сtg a = (4 * 5) / (5 * 3);

сtg a = (4 * 1) / (1 * 3);

сtg a = 4/3;

Ответ: tg a = 3/4 и сtg a = 4/3.