Let tan 30^{circ} and tan 15^{circ} be the ro | vedclass

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(A) Given that $	an 30^{\circ}$ and $	an 15^{\circ}$ are the roots of $x^2+ax+b=0$.We know that $	an 30^{\circ} = \frac{1}{\sqrt{3}}$.$	an 15^{\ci

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(A) Given that $	an 30^{\circ}$ and $	an 15^{\circ}$ are the roots of $x^2+ax+b=0$.We know that $	an 30^{\circ} = \frac{1}{\sqrt{3}}$.$	an 15^{\circ} = 	an(45^{\circ}-30^{\circ}) = \frac{	an 45^{\circ}-	an 30^{\circ}}{1+	an 45^{\circ}	an 30^{\circ}} = \frac{1-1/\sqrt{3}}{1+1/\sqrt{3}} = \frac{\sqrt{3}-1}{\sqrt{3}+1}$.Sum of roots: $-a = 	an 30^{\circ} + 	an 15^{\circ} = \frac{1}{\sqrt{3}} + \frac{\sqrt{3}-1}{\sqrt{3}+1} = \frac{\sqrt{3}+1+3-\sqrt{3}}{\sqrt{3}(\sqrt{3}+1)} = \frac{4}{\sqrt{3}(\sqrt{3}+1)}$.So,$a = -\frac{4}{3+\sqrt{3}}$.Product of roots: $b = 	an 30^{\circ} \cdot 	an 15^{\circ} = \frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}-1}{\sqrt{3}+1} = \frac{\sqrt{3}-1}{3+\sqrt{3}}$.Now,$1+a-b = 1 - \frac{4}{\sqrt{3}(\sqrt{3}+1)} - \frac{\sqrt{3}-1}{\sqrt{3}(\sqrt{3}+1)} = \frac{3+\sqrt{3}-4-\sqrt{3}+1}{\sqrt{3}(\sqrt{3}+1)} = \frac{0}{\sqrt{3}(\sqrt{3}+1)} = 0$.

Let $\tan 30^{\circ}$ and $\tan 15^{\circ}$ be the roots of the quadratic equation $x^2+ax+b=0$,then $1+a-b=$

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