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

Question

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

Dr. Abubacker · Accepted Answer

(b)
Given, $\sin{\frac{	heta}{2} = \sqrt{\frac{x - 1}{2x}}}$
$	herefore\ 	an{\frac{	heta}{2} = \sqrt{\frac{x - 1}{x + 1}}}$
$	herefore\ \ 	an{	heta =}\frac{2	an\frac{	heta}{2}}{1 - 	an^{2}\frac{	heta}{2}}$
$= \frac{2\sqrt{\frac{x - 1}{x + 1}}}{1 - \frac{x - 1}{x + 1}} = \frac{2\sqrt{\frac{x - 1}{x + 1}}}{\frac{2}{x + 1}} = \sqrt{x^{2} - 1}$
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Q. If $\theta$ is an acute angle and $\sin{\frac{\theta}{2} = \sqrt{\frac{x - 1}{2x}}}$, then $\tan\theta$ is equal to

Explanation

Reviewed by Dr. Abubacker

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