f.3 maths please~(15分)3條~~

1.若兩條直線L1:ax+3y-1=0和L2:x-6y+2=0平行,求a的值L1 : ax+3y-1=0 3y = -ax + 1y = (-a/3)x + 1/3L1 斜率 = -a/3;L2 : x-6y+2=06y = x + 2y = (1/6)x + 1/3L2 斜率 =1/6L1 // L2 :-a/3 = 1/6-a = 1/2a = - 1/2
2.投擲3顆骰子,求下列情況的概率
(a):擲得總和是18:P(6 , 6 , 6) = (1/6)(1/6)(1/6) = 1/216(b):擲得1個「3」P(3 , 非3 , 非3) + P(非3 , 3 , 非3) + P(非3 , 非3 , 3)= (1/6)(5/6)(5/6) + (5/6)(1/6)(5/6) + (5/6)(5/6)(1/6)= 3 * (1/6)(5/6)^2= 25/72(c):擲得2個「5」P( 5 , 5 , 非5) + P(5 , 非5 , 5) + P(非5 , 5 , 5)= 3 * (5/6)(1/6)^2= 5/72

3.已知tan⊕=sin⊕/cos⊕ 及 sin^2⊕+cos^2⊕=1,證明下列恆等式
(a)tan⊕/1+tan^2⊕=sin⊕cos⊕tan⊕/1+tan^2⊕= tan⊕ / [ 1+ (sin^2 ⊕) / (cos^2 ⊕) ]= tan⊕ / [ (sin^2 ⊕ + cos^2 ⊕) / (cos^2 ⊕) ]= tan⊕ / [1/(cos^2 ⊕)]= tan⊕ cos^2 ⊕= (sin⊕ / cos⊕) cos^2 ⊕= sin⊕cos⊕