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If the angle between the asymptotes of a hyperbola is $30^{\circ}$ then its eccentricity is
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The correct answer is:
$\sqrt{6}-\sqrt{2}$
Angle between asymptotes of hyperbola $=2 \sec ^{-1}(e)$
$\begin{aligned} & 30^{\circ}=2 \sec ^{-1} e \Rightarrow \sec ^{-1} e=15^{\circ} \\ & \therefore e=\sec 15^{\circ} \\ & e=\sqrt{6}-\sqrt{2}\end{aligned}$
$\begin{aligned} & 30^{\circ}=2 \sec ^{-1} e \Rightarrow \sec ^{-1} e=15^{\circ} \\ & \therefore e=\sec 15^{\circ} \\ & e=\sqrt{6}-\sqrt{2}\end{aligned}$
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