The larger of $\cos (\log heta)$ and $\log (\cos heta)$, if $e^{-\pi / 2}

Question

The larger of $\cos (\log 	heta)$ and $\log (\cos 	heta)$, if $e^{-\pi / 2} < 	heta < \pi / 2$ is

Aicharya · Accepted Answer

$ \begin{aligned} & 	ext { } \cos (\log 	heta), 	heta \in\left(e^{\frac{-\pi}{2}}, \frac{\pi}{2}ight) \ & e^{-\frac{\pi}{2}} < 	heta < \frac{\pi}{2} \ & \Rightarrow \quad-\frac{\pi}{2} < \log 	heta < \log \frac{\pi}{2} \leq \frac{\pi}{2} \end{aligned} $ So, $\cos (\log 	heta) \geq 0$ $ \log (\cos 	heta) \leq 0 $ $ 0 \leq \cos 	heta \leq 1 	ext { in } 	heta \in\left[e^{-\frac{\pi}{2}}, \frac{\pi}{2}ight] $ $ 	herefore \quad \log (\cos 	heta) \leq 0 $ Hence, $\cos (\log 	heta) \geq \log (\cos 	heta)$

The larger of $\cos (\log \theta)$ and $\log (\cos \theta)$, if $e^{-\pi / 2} < \theta < \pi / 2$ is

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