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

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
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

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

csc^2 x=cot^2 x +1，cot^2 x =1/tan^2 x
y=(tan^2 x-csc^2 x )/(tan^2 x+cot^2 x -1)
=(tan^2 x- cot^2 x -1)/(tan^2 x+cot^2 x -1)
=1 - 2cot^2 x /(tan^2 x+cot^2 x -1)
=1-2/（tan^4 x-tan^2 x +...

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csc^2 x=cot^2 x +1，cot^2 x =1/tan^2 x
y=(tan^2 x-csc^2 x )/(tan^2 x+cot^2 x -1)
=(tan^2 x- cot^2 x -1)/(tan^2 x+cot^2 x -1)
=1 - 2cot^2 x /(tan^2 x+cot^2 x -1)
=1-2/（tan^4 x-tan^2 x +1）
证毕
b）、y=1-2/（tan^4 x-tan^2 x +1）=1-2/[（tan^2x-1/2)^+3/4]
tan^2x≥0，所以（tan^2x-1/2)^+3/4≥3/4
当tan^2x=1/2时，ymin=1-2*4/3=-5/3
y值的范围【-5/3，1）

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