Mathematics - Quadratic Equation Question with Solution | TestHub

$\because \tan 15^{\circ }=\tan (45-30{)}^{\circ }=\frac{\sqrt{3}-1}{\sqrt{3}+1}$ $\therefore \tan 15^{\circ }+\tan 30^{\circ }=-p$ $\Rightarrow p=\frac{-4}{\sqrt{3}(\sqrt{3}+1)}$ $q=\tan 15^{\circ }\times \tan 30^{\circ }$ $=\frac{\sqrt{3}-1}{\sqrt{3}(\sqrt{3}+1)}\Rightarrow q=\frac{\sqrt{3}-1}{\sqrt{3}(\sqrt{3}+1)}$ $\therefore p\cdot q=\frac{-4}{\sqrt{3}(\sqrt{3}+1)}\times \frac{(\sqrt{3}-1)}{\sqrt{3}(\sqrt{3}+1)}$ $\Rightarrow p\cdot q=\frac{-4(\sqrt{3}-1)}{3(\sqrt{3}+1{)}^2}=\frac{10-6\sqrt{3}}{3}$ $\Rightarrow p\cdot q=\frac{10-6\sqrt{3}}{3}$ $\Rightarrow$ Option (2) is correct.