a.maths 可以幫幫忙嗎？不會做呀

1. cos ^2 ( 180 - A ) + cos ^2 ( 270 - A )
= cos^2 A+ sin^2 A as cos (180-A)= - cos A and cos (270-A)= - sin A, after suqaring, the negative sign can be omitted.
= 1

2. 1+tan^2 (270+A)
= 1+ tan^2 (90-A)
= 1+ cot^2 A [By the law tan a= 1/tan(90-a)= cot (90-a)]
= csc^2 A [as 1+cot^2 A =csc ^2 A]

3. 1 - cos A 1 + cos A
-------------- X ------------------
sin A 1 + cos A
1 - cos^2 A
= --------------------------------
sin A X (cos A+ 1)
sin^2 A
= ---------------------------
sinA X (1+cos A)
sin A
= --------------------
1 + cos A

4. 2(sin A+ cos A)= 開方3 + 1
Squaring both sides, we have
4(sin^2 A+cos^2 A +2sinAcosA)=4+2 x 開方3
8sinAcosA = 2 x 開方3
4sin2A = 2X 開方3
sin 2A= 開方3/2
2A= 兀/3 OR 2兀/3
A=兀/6 OR 兀/3
For this question, more info for the range of A is required to have a more detailed soln.

5. 6 tan ^2 A - 4 sin ^2 A = 1
6 (sec^2 A-1) -4(1-cos^2 A)= 1
6sec^2 A - 6 - 4 + 4cos^2 A=1
4cos^4 A - 11cos^2 A +6 =0
Solving the quadraic equation, we have
cos^2 A = 0.75 OR 2 (rejected)
cos A= 開方3 /2 OR -開方3 /2
A= 兀/6 5兀/6 7兀/6 11兀/6 (as 0=<2兀)

6. tan ^2 A = 2 + 4 cos ^2 A
sec^2 A- 1 = 2 + 4cos^2 A
4cos^4 A +3 cos^2 A -1=0
cos ^2 A= 0.25 OR -1 (rejected)
cos A = 0.5 OR -0.5
A= 兀/3 2兀/3 4兀/3 5兀/3 (as 0=<2兀)

7.I don't know what the question is asking, plz specify.

8. k/3=sinAcosA (product of roots) and k=3sinAcosA
sin A+cos A= 2/3 (sum of roots)
sin^2 A +2sinAcosA +cos^2 A=4/9 (Squaring both sides)
2sinAcosA =4/9 -1
sinAcosA = -5/18
Thus, k=3X(-5/18)
= - 5/6

useπ, √
don't use兀