A) \[\frac{\sqrt{1-{{x}^{2}}}}{x}\]
B) \[\frac{x}{1+{{x}^{2}}}\]
C) \[\frac{\sqrt{1+{{x}^{2}}}}{x}\]
D) \[\sqrt{1-{{x}^{2}}}\]
Correct Answer: A
Solution :
Let \[{{\cos }^{-1}}x=\theta .\] Then \[x=\cos \theta \] \[\Rightarrow \,\,\tan \theta =\sqrt{{{\sec }^{2}}\theta -1}\]\[=\sqrt{\frac{1}{{{x}^{2}}}-1}=\sqrt{\frac{1-{{x}^{2}}}{x}}\] \[\therefore \,\,\tan \,({{\cos }^{-1}}x)=\tan \theta =\frac{\sqrt{1-{{x}^{2}}}}{x}\].
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