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.