Question

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

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

Answer 1

Answer: (B)

$\sqrt{x^2-1}$

$\sin \frac{\theta }{2}=\sqrt{\frac{x-1}{2x}}$

$\Rightarrow {\sin }^2\frac{\theta }{2}=\frac{x-1}{2x}$

$\Rightarrow 1-{\cos }^2\frac{\theta }{2}=\frac{x-1}{2x}$

$\Rightarrow {\cos }^2\frac{\theta }{2}=1-\frac{x-1}{2x}$

$\Rightarrow {\cos }^2\frac{\theta }{2}=\frac{2x-x+1}{2x}$

$\Rightarrow {\cos }^2\frac{\theta }{2}=\frac{x+1}{2x}$

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

$\Rightarrow \sin \frac{\theta }{2}.\cos \frac{\theta }{2}=\sqrt{\frac{x-1}{2x}}.\sqrt{\frac{x+1}{2x}}$

$\Rightarrow \sin \frac{\theta }{2}.\cos \frac{\theta }{2}=\frac{1}{2x}\sqrt{x^2-1}$

$\Rightarrow 2\sin \frac{\theta }{2}.\cos \frac{\theta }{2}=\frac{1}{x}\sqrt{x^2-1}$

$\Rightarrow \sin \theta =\frac{1}{x}\sqrt{x^2-1}$

$\Rightarrow {\sin }^2\theta =\frac{x^2-1}{x^2}$

$\Rightarrow 1-{\cos }^2\theta =\frac{x^2-1}{x^2}$

$\Rightarrow {\cos }^2\theta =1-\frac{x^2-1}{x^2}$

$\Rightarrow {\cos }^2\theta =\frac{1}{x^2}$

$\Rightarrow \cos \theta =\frac{1}{x}$

$\therefore \tan \theta =\frac{\sin \theta }{\cos \theta }=\frac{\frac{\sqrt{x^2-1}}{x}}{\frac{1}{x}}=\frac{\sqrt{x^2-1}}{x}\times \frac{x}{1}$

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

Hence, the answer is $\sqrt{x^2-1}.$