If theta is an acute angle and sin frac{theta | vedclass

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(B) Given $\sin \frac{	heta}{2} = \sqrt{\frac{x - 1}{2x}}$.We know that $\cos 	heta = 1 - 2\sin^2 \frac{	heta}{2} = 1 - 2\left(\frac{x - 1}{2x}ig

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$\sqrt{x^2 - 1}$

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(B) Given $\sin \frac{	heta}{2} = \sqrt{\frac{x - 1}{2x}}$.We know that $\cos 	heta = 1 - 2\sin^2 \frac{	heta}{2} = 1 - 2\left(\frac{x - 1}{2x}ight) = 1 - \frac{x - 1}{x} = \frac{x - x + 1}{x} = \frac{1}{x}$.Also,$\sin 	heta = 2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}$.Since $\sin \frac{	heta}{2} = \sqrt{\frac{x - 1}{2x}}$,then $\cos \frac{	heta}{2} = \sqrt{1 - \sin^2 \frac{	heta}{2}} = \sqrt{1 - \frac{x - 1}{2x}} = \sqrt{\frac{2x - x + 1}{2x}} = \sqrt{\frac{x + 1}{2x}}$.Thus,$\sin 	heta = 2 \sqrt{\frac{x - 1}{2x}} \cdot \sqrt{\frac{x + 1}{2x}} = 2 \frac{\sqrt{x^2 - 1}}{2x} = \frac{\sqrt{x^2 - 1}}{x}$.Finally,$	an 	heta = \frac{\sin 	heta}{\cos 	heta} = \frac{\frac{\sqrt{x^2 - 1}}{x}}{\frac{1}{x}} = \sqrt{x^2 - 1}$.

If $\theta$ is an acute angle and $\sin \frac{\theta}{2} = \sqrt{\frac{x - 1}{2x}}$,then $\tan \theta$ is equal to

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