Cosine of an Integer

Solution as suggested by Sid Rana

$\begin{aligned} \tan \frac{\pi}{12} & = \tan \left(\frac{\pi}{4} - \frac{\pi}{6}\right) \\ & = \frac{1 - \frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}} \\ & = \frac{\sqrt{3}-1}{\sqrt{3}+1} \\ & = \sqrt{\frac{(\sqrt{3}-1)^2}{(\sqrt{3}+1)^2}} \\ & = \sqrt{\frac{4-2\sqrt{3}}{4+2\sqrt{3}}} \\ & = \sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}}} \end{aligned}$

$\implies a \times b = 2 \times 3 = \boxed{6}$

Previous solution

$\begin{aligned} \tan \left(\frac{\pi}{6} \right) & = \frac{1}{\sqrt{3}} \\ \implies \frac{2\tan \left(\frac{\pi}{12} \right)}{1-\tan^2 \left(\frac{\pi}{12} \right)} & = \frac{1}{\sqrt{3}} \\ \tan^2 \left(\frac{\pi}{12} \right) + 2 \sqrt{3} \tan \left(\frac{\pi}{12} \right) - 1 & = 0 \\ \implies \tan \left(\frac{\pi}{12} \right) & = \frac{-2\sqrt{3} \pm \sqrt{12+4}}{2} \\ & = 2 - \sqrt{3} \quad \quad \small \color{#3D99F6}{\text{Since } \tan \left(\frac{\pi}{12} \right) > 0, \, -2-\sqrt{3} \text{ is rejected.}} \\ & = \sqrt{\frac{(2-\sqrt{3})^2}{1}} \\ & = \sqrt{\frac{(2-\sqrt{3})^2}{\color{#3D99F6}{(2+\sqrt{3})(2-\sqrt{3})}}} \quad \quad \small \color{#3D99F6}{\text{Note that } (2+\sqrt{3})(2-\sqrt{3}) = 1} \\ \implies \tan \left(\frac{\pi}{12} \right) & = \sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}}} \end{aligned}$

$\implies a \times b = 2 \times 3 = \boxed{6}$