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Find the range of f(x)=sec(pi/4cos^2x)...

Find the range of `f(x)=sec(pi/4cos^2x)`

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The correct Answer is:
`[1,sqrt(2)]`
`f(x)="sec"((pi)/(4)"cos"^(2)x)`
We know that `0 le cos^(x) le 1,` i.e.,
`0 le (pi)/(4)cos^(2)x le (pi)/(4)`
For the above value of `theta =(pi)/(pi)/(4)cos^(2)x,sec x` is an increasing function.
` :. f(x) in [f(0),f((pi)/(4))] " or " f(x) in [1,sqrt(2)]`
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