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^2 \frac{	heta}{2} = 1 - \sin^2 \frac{	heta}{2} = 1 - \frac{x - 1}{2x}

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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 $\cos^2 \frac{	heta}{2} = 1 - \sin^2 \frac{	heta}{2} = 1 - \frac{x - 1}{2x} = \frac{2x - x + 1}{2x} = \frac{x + 1}{2x}$.Thus,$\cos \frac{	heta}{2} = \sqrt{\frac{x + 1}{2x}}$.Now,$	an \frac{	heta}{2} = \frac{\sin(	heta/2)}{\cos(	heta/2)} = \sqrt{\frac{x - 1}{x + 1}}$.Using the formula $	an 	heta = \frac{2 	an(	heta/2)}{1 - 	an^2(	heta/2)}$:$	an 	heta = \frac{2 \sqrt{\frac{x - 1}{x + 1}}}{1 - \frac{x - 1}{x + 1}} = \frac{2 \sqrt{\frac{x - 1}{x + 1}}}{\frac{x + 1 - x + 1}{x + 1}} = \frac{2 \sqrt{\frac{x - 1}{x + 1}}}{\frac{2}{x + 1}} = (x + 1) \sqrt{\frac{x - 1}{x + 1}} = \sqrt{(x + 1)^2 \cdot \frac{x - 1}{x + 1}} = \sqrt{(x + 1)(x - 1)} = \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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