Answer:
Given, \[f(x)=\log \,\,\cos \,x\] On differentiating both sides w.r.t. x, we get \[f'(x)=\frac{1}{\cos \,x}\cdot \frac{d}{dx}(\cos \,x)\] \[=\frac{1}{\cos \,x}\cdot (-\sin \,x)=-\tan \,x\] We know that, for \[x\in \left( 0,\,\,\frac{\pi }{2} \right),\,\,\tan \,x>0\] \[\therefore \] \[f'(x)=-\tan \,x<0\] Hence, \[f(x)\] is strictly decreasing. Hence proved.
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