A) \[\pi -{{\cos }^{-1}}\{\sqrt{1-{{x}^{2}}}\}\]
B) \[{{\tan }^{-1}}\left\{ \frac{x}{\sqrt{1-{{x}^{2}}}} \right\}\]
C) \[-{{\cot }^{-1}}\left\{ \frac{\sqrt{1-{{x}^{2}}}}{x} \right\}\]
D) \[\cos e{{c}^{-1}}x\]
Correct Answer: B
Solution :
\[\because \]\[-1<x<0,\]then\[-\frac{\pi }{2}<{{\sin }^{-1}}x<0\] Let\[{{\sin }^{-1}}x=\alpha \Rightarrow \sin \alpha =x\] Then, \[\tan \alpha =\frac{x}{\sqrt{1-{{x}^{2}}}}\] \[\Rightarrow \] \[\alpha ={{\tan }^{-1}}\left( \frac{x}{\sqrt{1-{{x}^{2}}}} \right)\] \[\therefore \] \[{{\sin }^{-1}}x={{\tan }^{-1}}\left( \frac{x}{\sqrt{1-{{x}^{2}}}} \right)\]
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