If α + β =

$\frac{\pi}{2}=90^{\circ}$

Let $x=\cos \alpha \cos \beta$

$\Rightarrow x=\frac{1}{2}[2 \cos \alpha \cos \beta]$

$\Rightarrow x=\frac{1}{2}[\cos (\alpha+\beta)+\cos (\alpha-\beta)]$

$\Rightarrow x=\frac{1}{2}\left[\cos (\alpha-\beta)+\cos 90^{\circ}\right]$

$\Rightarrow x=\frac{1}{2} \cos (\alpha-\beta)$

Now,

$-1 \leq \cos (\alpha-\beta) \leq 1$

$\Rightarrow-\frac{1}{2} \leq \frac{1}{2} \cos (\alpha-\beta) \leq \frac{1}{2}$

$\Rightarrow-\frac{1}{2} \leq x \leq \frac{1}{2}$

$\Rightarrow-\frac{1}{2} \leq \cos \alpha \cos \beta \leq \frac{1}{2}$

Hence, $\frac{1}{2}$ is the maximum value of $\cos \alpha \cos \beta$.