Ответ: a) π/2<t<π — 2-ая четвертьcost — » — «tgt и ctgt — » — «cost = -√(1-(4/5)²) = — √(1 — 16/25) = — √9/25 = -3/5tgt = sint / cost = 4/5 : (-3/5) = -4/3ctgt = 1/tgt = -3/4б) 0<t<π/2 — 1-ая четвертьcost , tgt, ctgt — » + «cost = √(1-(5/13)²) = √(1- 25/169) = √(144/169) = 12/13tgt= 5/13 : (12/13)=5/12ctgt = 12/5 в) -π/2<t<0 — 4-ая четвертьcost — » + «tgt, ctgt — » — «cost =  √(1-(-0.6)²) =  √(1-0.36) =  √0.64 =  0.8tgt = -0.6 / 0.8 =  — 6/8 =  — 3/4ctgt = — 4/3г) π<t<3π/2 — 3-я четвертьcost — » — «tgt, ctgt — » + «cost = — √(1-(-0.28)²) = — √0.9216 = — 0.96tgt = -0.28 / (-0.96) = 28/96 = 7/24ctgt = 24/7 = 3 ³/₇