A) \[\pm \sqrt{\frac{1+x}{{{x}^{2}}}},x\ne 0\]
B) \[\pm \sqrt{\frac{1+{{x}^{2}}}{{{x}^{2}}}},x\ne 0\]
C) \[\frac{-x}{\sqrt{1+{{x}^{2}}}}\]
D) \[\sqrt{\frac{1-{{x}^{2}}}{x}},x\ne 0\]
Correct Answer: D
Solution :
[d] \[\because \tan \left\{ {{\cos }^{-1}}(x) \right\}\] \[\tan \left\{ {{\tan }^{-1}}\left( \frac{\sqrt{1-{{x}^{2}}}}{x} \right) \right\}=\frac{\sqrt{1-{{x}^{2}}}}{x},x\ne 0\] Hence, option [d] is correct.
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