M2 MATHS F.4

1a)
18 + 12√2
= 18 + 2√72
= 6 + 2√(6*12) + 12
= √6 + √12
a = 6 , b = 12.
b)
1 / (18 + 12√2)
= 1 / (√6 + √12)^2
so the positive square root of it
= 1 / (√6 + √12)
= (√6 - √12) / (6 - 12)
= (√6 - 2√3) / ( - 6)
= (2√3 - √6) / 6

2)
tanA = cotB
sinA / cosA = cosB / sinB
cosA cosB - sinA sinB = 0
cos (A+B) = 0
(A+B) = π/2 or 3π/2 ,
Since tanA = cotB for 0 < A (or B) < π , so there are two possible :
Both A and B lie on the 1st quadrant , i.e. A + B < π/2 + π/2 = π ,
so A+B = π/2
or both lie on the 2nd quadrant , i.e. π/2 + π/2 = π < A + B < π + π = 2π ,
so A+B = 3π/2 , for example , A = B = 3π/4 ,
then
tan 3π/4 = cot 3π/4 = - 1 , were 0 < A or B =3π/4 < π ,
A+B = 3π/4 + 3π/4 is not = π/2 .
I think the question have some mistakes .
(or my solution have some mistakes.)


2010-07-13 16:39:00 補充：
Line 5 should be

= (√6 + √12)^2