If the angle between the asymptotes of a hype | vedclass

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(D) The angle between the asymptotes of a hyperbola is given by $2 \sec^{-1}(e)$.Given that the angle is $30^{\circ}$,we have:$2 \sec^{-1}(e) = 30^{\c

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$\sqrt{6}-\sqrt{2}$

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(D) The angle between the asymptotes of a hyperbola is given by $2 \sec^{-1}(e)$.Given that the angle is $30^{\circ}$,we have:$2 \sec^{-1}(e) = 30^{\circ}$$\sec^{-1}(e) = 15^{\circ}$$e = \sec(15^{\circ})$Since $\cos(15^{\circ}) = \cos(45^{\circ}-30^{\circ}) = \cos(45^{\circ}) \cos(30^{\circ}) + \sin(45^{\circ}) \sin(30^{\circ}) = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3}+1}{2\sqrt{2}}$.Therefore,$e = \frac{2\sqrt{2}}{\sqrt{3}+1} = \frac{2\sqrt{2}(\sqrt{3}-1)}{3-1} = \sqrt{6}-\sqrt{2}$.

If the angle between the asymptotes of a hyperbola is $30^{\circ}$,then its eccentricity is

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