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Find the local maximum and local minimum value of `f(x)=secx+logcos^2x ,0ltxlt2pi`

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`f(x)=sec x+2 log cos x`

Therefore, `f^{prime}(x)=sec x tan x-2 tan x=tan x(sec x-2)`

`f^{prime}(x)=0`

`Rightarrow tan x=0 text { or secx }=2 text { or } cos x=frac{1}{2}`

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