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(C) The given hyperbola is $x^2 - y^2 \operatorname{cosec}^2 	heta = 5$,which can be written as $\frac{x^2}{5} - \frac{y^2}{5 \sin^2 	heta} = 1$. It

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$\frac{\pi}{3}$

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(C) The given hyperbola is $x^2 - y^2 \operatorname{cosec}^2 	heta = 5$,which can be written as $\frac{x^2}{5} - \frac{y^2}{5 \sin^2 	heta} = 1$. Its eccentricity $e_h$ is given by $e_h = \sqrt{1 + \frac{5 \sin^2 	heta}{5}} = \sqrt{1 + \sin^2 	heta}$. The given ellipse is $x^2 \operatorname{cosec}^2 	heta + y^2 = 5$,which can be written as $\frac{x^2}{5 \sin^2 	heta} + \frac{y^2}{5} = 1$. Since $\sin^2 	heta Its eccentricity $e_c$ is given by $e_c = \sqrt{1 - \frac{5 \sin^2 	heta}{5}} = \sqrt{1 - \sin^2 	heta} = \cos 	heta$. Given $e_h = \sqrt{7} e_c$,we have: $\sqrt{1 + \sin^2 	heta} = \sqrt{7} \cos 	heta$. Squaring both sides: $1 + \sin^2 	heta = 7 \cos^2 	heta$. $1 + \sin^2 	heta = 7(1 - \sin^2 	heta)$. $1 + \sin^2 	heta = 7 - 7 \sin^2 	heta$. $8 \sin^2 	heta = 6$. $\sin^2 	heta = \frac{3}{4}$. Since $0 Therefore,$	heta = \frac{\pi}{3}$.

For $0 < \theta < \pi / 2$,if the eccentricity of the hyperbola $x^2 - y^2 \operatorname{cosec}^2 \theta = 5$ is $\sqrt{7}$ times the eccentricity of the ellipse $x^2 \operatorname{cosec}^2 \theta + y^2 = 5$,then the value of $\theta$ is:

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