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 :
If \[{{\cos }^{-1}}\left( \frac{1}{x} \right)=\theta \] \[\frac{1}{x}=\cos \theta \] Now, \[\tan \theta =\frac{\sin \theta }{\cos \theta }\] \[=\frac{\sqrt{1-{{\cos }^{2}}\theta }}{\cos \theta }=\frac{\sqrt{1-\frac{1}{{{x}^{2}}}}}{\frac{1}{x}}\] \[=\sqrt{{{x}^{2}}-1}\]
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