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

Toolbox:
• A function $f(x)$ is said to be a strictly increasing function on $(a,b)$ if $x_1 < x_2\Rightarrow f(x_1) < f(x_2)$ for all $x_1,x_2\in (a,b)$
• If $x_1 < x_2\Rightarrow f(x_1) > f(x_2)$ for all $x_1,x_2\in (a,b)$ then $f(x)$ is said to be strictly decreasing on $(a,b)$
• A function $f(x)$ is said to be increasing on $[a,b]$ if it is increasing (decreasing) on $(a,b)$ and it is increasing (decreasing) at $x=a$ and $x=b$.
• The necessary sufficient condition for a differentiable function defined on $(a,b)$ to be strictly increasing on $(a,b)$ is that $f'(x) > 0$ for all $x\in (a,b)$
• The necessary sufficient condition for a differentiable function defined on $(a,b)$ to be strictly decreasing on $(a,b)$ is that $f'(x) < 0$ for all $x\in (a,b)$
Step 1:
Let $f(x)=\log|\cos x|$
Differentiating w.r.t $x$ we get,
$f'(x)=\large\frac{1}{\cos x}$$(-\sin x)=-|\tan x| Step 2: Consider the interval (0,\large\frac{\pi}{2}) \tan x > 0 But -\tan x<0 Therefore f'(x) < 0 on (0,\large\frac{\pi}{2}) Step 3: Consider the interval (\large\frac{\pi}{2},$$\pi)$
$\tan x <0\Rightarrow -(-\tan x)=\tan x >0$
Therefore $f'(x) >0$ on $(\large\frac{\pi}{2},$$\pi) Hence the function \log|\cos x| is strictly decreasing on (0,\large\frac{\pi}{2}) and strictly increasing on (\large\frac{\pi}{2},$$\pi)$