1..
f(x)=[1-sin(2x)]/[1-cos^2(π/2 -x)]
=[sin^2(x)+cos^2(x)-2sinxcosx]/[1-sin^2(x)]
=[sin^2(x)+cos^2(x)-2sinxcosx]/(cosx)^2
=(tanx)^2-2tanx +1
=(tanx-1)^2
f(α)=(tanα -1)^2=(-2-1)^2=9
2.
f(x)有意義,cosx≠0,cotx有意義,sinx≠0,綜上,得sinx≠0且cosx≠0
x≠kπ/2 (k∈Z),函數(shù)定義域{x|x≠kπ/2 ,k∈Z}
f(x)=cotxf(x)
=cotx[(tanx)^2 -2tanx +1]
=tanx -2 + 1/tanx
由均值不等式得tanx +1/tanx≥2或tanx +1/tanx≤-2
y≥0或y≤-4
函數(shù)的值域為(-∞,-4]U[0,+∞)