Trigonometric Identities Question 345
Question: If $ \theta $ is an acute angle and $ \sin \frac{\theta }{2}=\sqrt{\frac{x-1}{2x}} $ , then $ \tan \theta $ is equal to
[Orissa JEE 2005]
Options:
A) $ x^{2}-1 $
B) $ \sqrt{x^{2}-1} $
C) $ \sqrt{x^{2}+1} $
D) $ x^{2}+1 $
Show Answer
Answer:
Correct Answer: B
Solution:
$ \tan \theta =\frac{\sin \theta }{\cos \theta } $ $ \tan \theta =\frac{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}{1-2{{\sin }^{2}}\frac{\theta }{2}}=\frac{2\tan \frac{\theta }{2}}{1-{{\tan }^{2}}\frac{\theta }{2}} $ $ [ \begin{aligned} & Usingsin\frac{\theta }{2}=\sqrt{\frac{x-1}{2x}} \\ & \therefore \cos \frac{\theta }{2}=\sqrt{1-{{\sin }^{2}}\frac{\theta }{2}}=\sqrt{\frac{x+1}{2x}}\text{ and},tan\frac{\theta }{2}=\frac{\sqrt{x-1}}{\sqrt{x+1}} \\ \end{aligned} ] $
$ \therefore ,\tan \theta =\sqrt{x^{2}-1} $ .
BETA
