Question

If $\alpha$ is an acute anlge and $\sin \left(\frac{\alpha }{2}\right)=\sqrt{\frac{x-1}{2x}}$, then $\tan \alpha$ is

• A. $\sqrt{\frac{x-1}{x+1}}$
• B. $\frac{\sqrt{x-1}}{x+1}$
• C. $\sqrt{x^2-1}$
• D. $\sqrt{x^2+1}$

Answer 1

The correct option is C $\sqrt{x^2-1}$

$\sin \frac{\alpha }{2}=\sqrt{\frac{x-1}{2x}}$
${\sin }^2\frac{\alpha }{2}=\frac{x-1}{2x}$ ...... $\left(i\right)$
$1-{\cos }^2\left(\frac{\alpha }{2}\right)=\frac{x-1}{2x}[\because {\sin }^{2}x+{\cos }^{2}x=1]$
${\cos }^2\frac{\alpha }{2}=1-\left(\frac{x-1}{2x}\right)$
${\cos }^2\frac{\alpha }{2}=\frac{2x-x+1}{2x}$
${\cos }^2\frac{\alpha }{2}=\frac{x+1}{2x}$ ...... $\left(ii\right)$
From $\left(i\right)$ and $\left(ii\right)$, we get
${\tan }^2\frac{\alpha }{2}=\frac{{\sin }^2\frac{\alpha }{2}}{{\cos }^2\frac{\alpha }{2}}=\frac{x-1}{x+1}$
$\tan \frac{\alpha }{2}=\sqrt{\frac{x-1}{x+1}}$
Using formula, $\tan 2\theta =\frac{2\tan \theta }{1-{\tan }^2\theta }$
$\tan \alpha =\frac{2\tan \frac{\alpha }{2}}{1-{\tan }^2\frac{\alpha }{2}}$
$\therefore \tan \alpha =\frac{2\sqrt{\frac{x-1}{x+1}}}{1-\left[\frac{x-1}{x+1}\right]}$
$\tan \alpha =\frac{2\sqrt{\frac{x-1}{x+1}}}{\frac{x+1-x+1}{x+1}}$
$\tan \alpha =\frac{2(x+1)}{2}\sqrt{\frac{x-1}{x+1}}$
$\tan \alpha =\sqrt{(x+1)(x-1)}$
$\tan \alpha =\sqrt{x^2-1}$