If tan 15^{circ} and tan 30^{circ} are the ro | vedclass

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(B) Given that $	an 15^{\circ}$ and $	an 30^{\circ}$ are the roots of $x^2+px+q=0$. From the relation between roots and coefficients: $-p = 	an 15^

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$\frac{10-6\sqrt{3}}{3}$

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(B) Given that $	an 15^{\circ}$ and $	an 30^{\circ}$ are the roots of $x^2+px+q=0$. From the relation between roots and coefficients: $-p = 	an 15^{\circ} + 	an 30^{\circ}$ and $q = 	an 15^{\circ} 	an 30^{\circ}$. We know $	an 15^{\circ} = 	an(45^{\circ}-30^{\circ}) = \frac{	an 45^{\circ}-	an 30^{\circ}}{1+	an 45^{\circ}	an 30^{\circ}} = \frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}} = \frac{\sqrt{3}-1}{\sqrt{3}+1} = 2-\sqrt{3}$. Now,$q = (2-\sqrt{3}) 	imes \frac{1}{\sqrt{3}} = \frac{2\sqrt{3}-3}{3}$. And $-p = (2-\sqrt{3}) + \frac{1}{\sqrt{3}} = \frac{2\sqrt{3}-3+1}{\sqrt{3}} = \frac{2\sqrt{3}-2}{\sqrt{3}}$. So,$p = \frac{2-2\sqrt{3}}{\sqrt{3}}$. Then $pq = \left(\frac{2-2\sqrt{3}}{\sqrt{3}}ight) \left(\frac{2\sqrt{3}-3}{3}ight) = \frac{4\sqrt{3}-6-12+6\sqrt{3}}{3\sqrt{3}} = \frac{10\sqrt{3}-18}{3\sqrt{3}} = \frac{10}{3} - \frac{18}{3\sqrt{3}} = \frac{10}{3} - 2\sqrt{3} = \frac{10-6\sqrt{3}}{3}$.

If $\tan 15^{\circ}$ and $\tan 30^{\circ}$ are the roots of the equation $x^2+px+q=0$,then $pq=$

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