Question

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

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

Answer 1

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

Given: $\theta$ is an acute angle and $\sin \frac{\theta }{2}=\sqrt{\frac{x-1}{2x}}$
To find: $\tan \theta$
Now, from the sub multiple angle formulae of $\tan \theta$
i.e. $\tan \theta =\frac{2\tan \frac{\theta }{2}}{1-{\tan }^2\frac{\theta }{2}}$

Now, $0<\theta <\frac{\pi }{2}\Rightarrow 0<\frac{\theta }{2}<\frac{\pi }{4}$
Hence, all trigonometric ratios will be positive.
Now, using the identity ${\sin }^{2}A+{\cos }^{2}A=1$
$\text{we can say,}\cos \frac{\theta }{2}=\sqrt{1-{\sin }^2\frac{\theta }{2}}$
$\Rightarrow \cos \frac{\theta }{2}=\sqrt{1-\frac{x-1}{2x}}$
$\Rightarrow \cos \frac{\theta }{2}=\sqrt{\frac{x+1}{2x}}$
$\Rightarrow \tan \frac{\theta }{2}=\frac{\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}}=\sqrt{\frac{x-1}{x+1}}$
Substituting in the formulae for submultiple angle, we get:
$\tan \theta =\frac{2\sqrt{\frac{x-1}{x+1}}}{1-\frac{x-1}{x+1}}$

$\Rightarrow \tan \theta =\frac{2\sqrt{\frac{x-1}{x+1}}}{\frac{2}{x+1}}$

$\Rightarrow \tan \theta =\sqrt{\frac{x-1}{x+1}\times (x+1{)}^2}=\sqrt{x^2-1}$
$\therefore \tan \theta =\sqrt{x^2-1}$