Помогите с 1,2,5.Предмет алгебра. ОЧЕНЬ СРОЧНО!!

ко второму заданию

5/
(ctg a+ctg b)/(sin a+sin b)=(cos a/sin a+cos b/sin b)/(sin a+sin b)=
=((cos a·sin b+cos b·sin a)/(sin a·sin b))/(sin a+sin b)=
=sin(a+b)/(sin a·sin b·(sin a+sin b))=1/(sin a·sin b)
ч т д

(1) [18 1/6 - (3,06 : 7 1/2 + 3 2/5 * 0,38)] : (19 - 2 3/8 * 5 1/3)
1) 3,06 : 7 1/2 = 3 3/50 : 7 1/2 = 153/50 : 15/2 = (153 * 2) / (50 * 15) = 51/125;
2) 3 2/5 * 0,38 = 17/5 * 19/50 = 323/250;
3) 51/125 + 323/250 = (102 + 323) / 250 = 425/250 = 17/10;
4) 18 1/6 - 17/10 = 109/6 - 17/10 = (545 - 51)/30 =494/30 = 247/15
5) 2 3/8 * 5 1/3 = 19/8 * 16/3 = 38/3
6) 19 - 38/3 = (57 - 38)/3 = 19/3
7) 247/15 : 19/3 = (247 * 3) / (15 * 19) = 13/5 = 2,6
Ответ: [18 1/6 - (3,06 : 7 1/2 + 3 2/5 * 0,38)] : (19 - 2 3/8 * 5 1/3) = 2,6

(2) 1/(5+2*корень (6))+1/(5-2*корень (6))=
=(5-2*корень (6)+5+2*корень (6))/[(5+2*корень (6))*(5-2*корень (6))]=
=10/(25-4*6)=10/(25-24)=10/1=10;

(3) (ctgA+ctgB)/sin(A+B)=
=(cosA/sinA+cosB/sinB)/sin(A+B)=
=[(cosA*sinB+cosB*sinA)/(sinA*sinB)]/sin(A+B)=
=[sin(A+B)/(sinA*sinB)]/sin(A+B)=1/(sinA*sinB)
1/(sinA*sinB)=1/(sinA*sinB) - тождество доказано