Prove that the function f given by `f (x) = log cos x` is strictly decreasing on `(0,pi/2)`and strictly increasing on `(pi/2,pi)`

Explanation

given `f(x)=logxcosx`

We have to prove `f(x)` is strictly increasing on `(-pi/2,0)` and strictly decreasing on `(0.pi/2)`

Now `f'(x)=1/cosx(-sinx)impliesf'(x)=-tanx`

Case I - Strictly increasing

`f'(x)>0implies-tanx>0impliestanx<0`

`x∈(-pi/2,0)or (pi/2,pi)`

Case II - Strictly decreasing

`f'(x)<0 implies-tanx<0impliestanx>0`

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