设y=(tan^2 x-csc^2 x )/(tan^2 x+cot^2 x -1) (a)证明y=1- 2/(tan^4 x-tan^2 x +1) (b)求y值的范围

1.
y=(tan^2 x-csc^2 x)/(tan^2 x+cot^2 x -1)
=(sin^2 x/cos^2 x-1/sin^2 x)/(sin^2 x/cos^2 x+cos^2 x/sin^2 x -1)
=(sin^4 x-cos^2 x)/(sin^4 x+cos^4 x-sin^2 xcos^2 x)
=(sin^4 x-cos^2 x)/[(sin^2 x+cos^2 x)^2-3sin^2 xcos^2 x]
=(sin^4 x+sin^2 x-1)/[1-3sin^2 xcos^2 x]
=[(sin^2 x+1)(sin^2 x-1)+sin^2 x]/[1-3sin^2 xcos^2 x]
=(-cos^2 xsin^2 x-cos^2 x+sin^2 x)/[1-3sin^2 xcos^2 x]
=1+(2cos^2 xsin^2 x-cos^2 x+sin^2 x-1)/[1-3sin^2 xcos^2 x]
=1+2(cos^2 xsin^2 x-cos^2 x)/[1-3sin^2 xcos^2 x]
=1-2cos^4 x/[1-3sin^2 xcos^2 x]
=1-2/[1/cos^4 x-3sin^2 x/cos^2 x]
=1-2/[(cos^2 x+sin^2 x)^2/cos^4 x-3sin^2 x/cos^2 x]
=1-2/[1+2cos^2 xsin^2 x/cos^4 x+sin^4 x/cos^4 x-3sin^2 x/cos^2 x]
=1-2/[1+tan^4 x-sin^2 x/cos^2 x]
=1-2/(tan^4 x-tan^2 x+1)
2.
y=1-2/(tan^4 x-tan^2 x+1)
tan^4 x-tan^2 x+1=2/(1-y)
(tan^2 x-1/2)^2+3/4=2/(1-y)
tan^2 x≥0
tan^2 x-1/2≥-1/2
当-1/2≤tan^2 x-1/2≤1/2时,
0≤(tan^2 x-1/2)^2≤1/4
3/4≤(tan^2 x-1/2)^2+3/4≤1
3/4≤2/(1-y)≤1,此时y＜1
1≤(1-y)/2≤4/3
-5/3≤y≤-1
所以-5/3≤y＜1；
当tan^2 x-1/2≥1/2时,
(tan^2 x-1/2)^2+3/4≥1
2/(1-y)≥1,此时y＜1
y≥-1
-1≤y＜1.
综上所述-5/3≤y＜1