thz...A.maths---urgent

[sec^2(π- x)] / {3cot^2 [(π/2) + x] -7} = 1
[sec^2(π- x)] ={3cot^2 [(π/2) + x] -7}
[1/cos^2(π- x)] ={3/tan^2 [(π/2) + x] -7}
since cos(π- x)=-cosx, sin(π- x)=sinx, tan[(π/2) + x]=-cotx
[1/cos^2(π- x)] ={3/tan^2 [(π/2) + x] -7}
1/cos^2x={3/cot^2x -7}
1/cos^2x={3sin^2x/cos^2x -7}
1=3sin^2x-7cos^2x
1=3(1-cos^2x)-7cos^2x
1=3-10cos^2x
cos^2x=1/5
cosx=-squareroot[1/5]


2007-01-27 20:58:15 補充：
it is because π/2<=x<=πcosx is negative

[sec^2(n - x)] / {3cot^2 [(n/2) + x] -7} = 1
sec^2 x=3tan^2 x-７
tan^2 x+1=3tan^2 x-7
2tan^2 x=8
tan^2 x=4
tanx= -2(Because n/2＜x