Question

tg a - ?
cos a = 5√29 /29 и a принадлежит (3п/2; 2п)

Answer 1

α∈(3π/2;2π)- IVчетверь⇒sinα<0; tgα<0
sin²α+cos²α=1
sin²α=1-25/29=4/29⇒sinα=-2√29/29
tgα=sinα/cosα=-2√29/29÷5√29/29=-2/5=-0,4