Question
Prove that the function f given by $\text{f}(\text{x})=\log\cos\text{x}$ is strictly increasing on $\Big(-\frac{\pi}{2},0\Big)$ and strictly decreasing on $\Big(0,\frac{\pi}{2}\Big).$

Answer

We have,
$\text{f}(\text{x})=\log\cos\text{x}$
$\therefore\ \text{f}'(\text{x})=\frac{1}{\cos\text{x}}(-\sin\text{x})=-\tan\text{x}$
In interval $\Big(0,\frac{\pi}{2}\Big),\tan\text{x}>0\Rightarrow-\tan\text{x}<0.$
$\therefore\text{f}'(\text{x})<0\text{ on }\Big(0,\frac{\pi}{2}\Big)$
$\therefore$ f is strictly decreasing on $\Big(0,\frac{\pi}{2}\Big).$
In interval $\Big(\frac{\pi}{2},\pi\Big),\tan\text{x}<0\Rightarrow-\tan\text{x}>0.$
$\therefore\text{f}'(\text{x})>0\text{ on }\Big(\frac{\pi}{2},\pi\Big)$
$\therefore$ f is strictly increasing on $\Big(-\frac{\pi}{2},0\Big).$

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