已知函数f(x)=sin^2x+csc^2x+cos^2x+sec^2x+tan^2x+7cot^2x,x∈(0,π/4

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已知函数f(x)=sin^2x+csc^2x+cos^2x+sec^2x+tan^2x+7cot^2x,x∈(0,π/4]求函数f(x)的最小值是多少着急呢

利用公式sec^2x=1+ tan^2x,csc^2x=1+ cot^2x,sin^2x+cos^2x=1
f(x)=sin^2x+csc^2x+cos^2x+sec^2x+tan^2x+7cot^2x
= sin^2x+cos^2x+tan^2x+7cot^2x+sec^2x+csc^2x
=1+ tan^2x+7cot^2x+1+ tan^2x+1+ cot^2x
=3+2 tan^2x+8cot^2x
=3+2 tan^2x+8/ tan^2x
设tan^2x=t,x∈(0,π/4]
则t∈(0,1].
2 tan^2x+8/ tan^2x=2t+8/t,该函数在(0,2]上递减,
所以2t+8/t的最小值是2+8=10.
∴函数f（x）的最小值是13.

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