Prove that the function $f$ given by $f\left( x \right) = \log \cos x$ is strictly decreasing on $\left( {0,\cfrac{\pi }{2}} \right)$ and strictly increasing on $\left( {\cfrac{\pi }{2},\pi } \right)$.

We have, $f(x) = \log (\cos x)$ …(i)

Differentiating the given function ie. (i) wr.t. $x$, we get $f'(x) = \cfrac{1}{{\cos x}}( - \sin x) = - \tan x$

As $\tan x > 0, - \tan x < 0$ for all $x \in \left( {0,\cfrac{\pi }{2}} \right)$ and
$\tan x < 0 \Rightarrow - \tan x > 0$ for all $x \in \left( {\cfrac{\pi }{2},\pi } \right)$

Therefore, $f(x) < 0$ for all $x \in \left( {0,\cfrac{\pi }{2}} \right)$ and $f(x) > 0$ for all $x \in \left( {\cfrac{\pi }{2},\pi } \right)$.

Hence, $f(x)$ is strictly decreasing on $\left( {0,\cfrac{\pi }{2}} \right)$ and strictly increasing on $\left( {\cfrac{\pi }{2},\pi } \right)$