A) \[{{x}^{2}}-1\]
B) \[\sqrt{{{x}^{2}}-1}\]
C) \[\sqrt{{{x}^{2}}+1}\]
D) \[{{x}^{2}}+1\]
Correct Answer: B
Solution :
\[\tan \theta =\frac{\sin \theta }{\cos \theta }\] \[\tan \theta =\frac{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}{1-2{{\sin }^{2}}\frac{\theta }{2}}=\frac{2\tan \frac{\theta }{2}}{1-{{\tan }^{2}}\frac{\theta }{2}}\] \[\left[ \begin{align} & \text{Using}\,\,\text{sin}\frac{\theta }{\text{2}}=\sqrt{\frac{x-1}{2x}} \\ & \therefore \,\,\cos \frac{\theta }{2}=\sqrt{1-{{\sin }^{2}}\frac{\theta }{2}}=\sqrt{\frac{x+1}{2x}}\text{ and}\,\text{tan}\frac{\theta }{\text{2}}=\frac{\sqrt{x-1}}{\sqrt{x+1}} \\ \end{align} \right]\] \[\therefore \,\,\,\tan \theta =\sqrt{{{x}^{2}}-1}\].
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