Which one is greater $\cos(\ln \theta)$ and $\ln(\cos \theta),$ Where $e^{-\frac{\pi}{2}}<\theta<\frac{\pi}{2}$ 
Which one is greater $\cos(\ln \theta)$ and $\ln(\cos \theta),$ Where $\displaystyle e^{-\frac{\pi}{2}}<\theta<\frac{\pi}{2}$

$\bf{My\; Try::}$ Given $\displaystyle e^{-\frac{\pi}{2}}<\theta<\frac{\pi}{2}.$ So $0<\cos \theta <1\;$
So $\displaystyle \ln(\cos \theta )<0$. Now $\displaystyle -\frac{\pi}{2}<\ln \theta < \ln \left(\frac{\pi}{2}\right)$
Now How can i solve it after that, Help required, Thanks
 A: Of course $\cos(\ln(x))>\ln(\cos(x))$ because in given range of $x$, $\cos(\ln(x))$ is positive as $\alpha(=\ln(x)) \in (-\frac{\pi}{2},\frac{\pi}{2})$ and we know that $\cos(\alpha)>0$ in this range of $\alpha$ and $\ln(\cos(x))$ is negative as you know $\ln(y)<0$ when $y<1$ .
A: Let $f(\theta) = \cos (\ln \theta)-\ln(\cos \theta)\;,$ Then $\displaystyle f'(\theta) = -\frac{\sin (\ln \theta)}{\theta}+\tan \theta$
Given $\displaystyle 0<e^{-\frac{\pi}{2}}<\theta <\frac{\pi}{2}\;,$ Then $\tan \theta >0$
and $\displaystyle -\frac{\pi}{2}<\ln (\theta)<\ln \left(\frac{\pi}{2}\right)<0$. So we get $\sin \left(\ln \theta \right)<0$
So we get $\displaystyle f'(\theta) =-\frac{\sin (\ln \theta)}{\theta}+\tan \theta>0$
Means $f(\theta)$ is strictly increasing function in $\displaystyle \theta \in \left(e^{-\frac{\pi}{2}},\frac{\pi}{2}\right)$
So we get $$\displaystyle f(\theta)>f\left(e^{-\frac{\pi}{2}}\right)\Rightarrow \cos(\ln \theta)-\ln(\cos \theta)>0-(\bf{-ve\; value}).$$
So $$\cos(\ln \theta)>\ln(\cos \theta).$$
