A) \[\frac{1}{\sqrt{{{x}^{2}}-1}}\]
B) \[\sqrt{{{x}^{2}}+1}\]
C) \[\sqrt{1-{{x}^{2}}}\]
D) \[\sqrt{{{x}^{2}}-1}\]
Correct Answer: D
Solution :
Given that \[{{\cos }^{-1}}\left( \frac{1}{x} \right)=\theta \,\,\Rightarrow \,\,\cos \theta =\frac{1}{x}\] Now, \[\tan \theta =\frac{\sin \theta }{\cos \theta }=\frac{\sqrt{1-{{(1/x)}^{2}}}}{1/x}=\sqrt{{{x}^{2}}-1}\]
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