Помогите пожалуйста с тригонометрией 10 класс, задание во вложении.

1а)sin(2π+π/6)=sinπ/6=1/2б)-tg(2π-π/6)=tgπ/6=√3/3в)cosπ+ctg(π+π/3)=-1+√3/3г)tgπ/4*(-ctgπ/4)+cos3π/2*sinπ/2=-1+0*1=-1д)sin(360+45)+cos(180+45)*tg(180+45)=sin45-cos45*tg45=√2/2-√2/2*1=02sin²t+cos²t/(ctgt*tgt)=sin²t+cos²t=131)cost=1/2t=+-π/3+2πn,n∈z2)cos(π/2+t)=-√3/2-sint=-√3/2sint=√3/2t=(-1)^n*π/3+πn,n∈z4cost=-4/5sint=-√(1-cos²t)=-√(1-16/25)=-√(9/25)=-3/5tgt=sint/cost=-3/5:(-4/5)=3/5*5/4=3/4ctgt=1/tgt=1:3/4=4/3

1.а) sin13π/6 = sin (2π+π/6 ) = sin π/6 =1/2.б) tq(-11π/6) = - tq11π/6 = - tq(2π -π/6 ) = tq π/6= √3 / 3.в) cos4π +ctq 4π/3 = 1 +ctq(π +π/3 ) =  1 +ctq π/3  =1 + √3 / 3.г) tq π/4 *ctq(- π/4) +cos(3π/2)* sinπ/2 =  -tq π/4 *ctq π/4 +0* sinπ/2  = -1.д) sin405° +cos225°*tq225° = sin(360°+45°) +sin (180°+ 45°)  =sin45° -sin45°=0.2.а) sin²t  - cos²t / ( ctqt *tq(-t) )=sin²t  - cos²t /(- ctqt *tqt  ) =.sin²t  + cos²t  =1. б)3.а) cost   =1 / 2 . ⇒ t = ± π/3 +2πn , n∈Z. б) cos(π/2 +t ) = -√3 /2   ⇒ - sin t  = -√3 /2 ⇔ sin t  = √3 /2 ⇒ t =(-1)^ n * π/3 +πn , n∈Z. 4.  cost  =- 4/5 ; π < t < 3π/2 .---------- sint - ?  ; tqt ? ;  ctqt - ? sint  = -√(1-cos²t) = -√(1-(4/5)= -3 /5 ; tqt =  sint  /cost  =(-3/5)/(-4/5) =3/4.ctqt  =1/tqt = 4/3 .5.sin(2π -t) -cos(3π/2 +t)  +1 = 0 ;-sint  -sint  +1 =0 ⇔sint =1/2⇒   t = (-1)^n  *( π/6) + πn ,n ∈Z.* * *   { π/6 + 2πn  ;  (π -π/6) +2πn  } ⇔     { π/6 + 2πn  ;  5π /6 +2πn  }  * * *