If tan frac{theta}{2} = operatorname{cosec} t | vedclass

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(B) Given,$	an \frac{	heta}{2} = \operatorname{cosec} 	heta - \sin 	heta$ $	an \frac{	heta}{2} = \frac{1}{\sin 	heta} - \sin 	heta = \frac{1 -

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$-2 + \sqrt{5}$

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(B) Given,$	an \frac{	heta}{2} = \operatorname{cosec} 	heta - \sin 	heta$ $	an \frac{	heta}{2} = \frac{1}{\sin 	heta} - \sin 	heta = \frac{1 - \sin^2 	heta}{\sin 	heta} = \frac{\cos^2 	heta}{\sin 	heta}$ Using half-angle formulas $\cos 	heta = \frac{1 - 	an^2 \frac{	heta}{2}}{1 + 	an^2 \frac{	heta}{2}}$ and $\sin 	heta = \frac{2 	an \frac{	heta}{2}}{1 + 	an^2 \frac{	heta}{2}}$: $	an \frac{	heta}{2} = \frac{(\frac{1 - 	an^2 \frac{	heta}{2}}{1 + 	an^2 \frac{	heta}{2}})^2}{\frac{2 	an \frac{	heta}{2}}{1 + 	an^2 \frac{	heta}{2}}} = \frac{(1 - 	an^2 \frac{	heta}{2})^2}{2 	an \frac{	heta}{2} (1 + 	an^2 \frac{	heta}{2})}$ Let $x = 	an^2 \frac{	heta}{2}$. Then $\sqrt{x} = \frac{(1 - x)^2}{2 \sqrt{x} (1 + x)}$ $2x(1 + x) = (1 - x)^2$ $2x + 2x^2 = 1 - 2x + x^2$ $x^2 + 4x - 1 = 0$ Solving for $x$ using the quadratic formula: $x = \frac{-4 \pm \sqrt{16 - 4(1)(-1)}}{2} = \frac{-4 \pm \sqrt{20}}{2} = -2 \pm \sqrt{5}$ Since $x = 	an^2 \frac{	heta}{2} > 0$,we have $x = -2 + \sqrt{5}$.

If $\tan \frac{\theta}{2} = \operatorname{cosec} \theta - \sin \theta$,then $\tan^2 \frac{\theta}{2} =$

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