задачу по комбинаторике (номер 1). Подробно с решением

A) π/2<t<π - 2-ая четверть
cost - " - "
tgt и ctgt - " - "
cost = -√(1-(4/5)²) = - √(1 - 16/25) = - √9/25 = -3/5
tgt = sint / cost = 4/5 : (-3/5) = -4/3
ctgt = 1/tgt = -3/4

б) 0<t<π/2 - 1-ая четверть
cost , tgt, ctgt - " + "
cost = √(1-(5/13)²) = √(1- 25/169) = √(144/169) = 12/13
tgt= 5/13 : (12/13)=5/12
ctgt = 12/5 

в) -π/2<t<0 - 4-ая четверть
cost - " + "
tgt, ctgt - " - "
cost =  √(1-(-0.6)²) =  √(1-0.36) =  √0.64 =  0.8
tgt = -0.6 / 0.8 =  - 6/8 =  - 3/4
ctgt = - 4/3

г) π<t<3π/2 - 3-я четверть
cost - " - "
tgt, ctgt - " + "
cost = - √(1-(-0.28)²) = - √0.9216 = - 0.96
tgt = -0.28 / (-0.96) = 28/96 = 7/24
ctgt = 24/7 = 3 ³/₇

A) π/2<t<π - 2-ая четверть
cost - " - "
tgt и ctgt - " - "
cost = -√(1-(4/5)²) = - √(1 - 16/25) = - √9/25 = -3/5
tgt = sint / cost = 4/5 : (-3/5) = -4/3
ctgt = 1/tgt = -3/4

б) 0<t<π/2 - 1-ая четверть
cost , tgt, ctgt - " + "
cost = √(1-(5/13)²) = √(1- 25/169) = √(144/169) = 12/13
tgt= 5/13 : (12/13)=5/12
ctgt = 12/5 

в) -π/2<t<0 - 4-ая четверть
cost - " + "
tgt, ctgt - " - "
cost =  √(1-(-0.6)²) =  √(1-0.36) =  √0.64 =  0.8
tgt = -0.6 / 0.8 =  - 6/8 =  - 3/4
ctgt = - 4/3

г) π<t<3π/2 - 3-я четверть
cost - " - "
tgt, ctgt - " + "
cost = - √(1-(-0.28)²) = - √0.9216 = - 0.96
tgt = -0.28 / (-0.96) = 28/96 = 7/24
ctgt = 24/7 = 3 ³/₇