A) \[x\]
B) \[\frac{1}{x}\]
C) \[1\]
D) \[0\]
Correct Answer: A
Solution :
[a] Let \[{{\cos }^{-1}}x=\theta \Rightarrow x=\cos \theta \] or \[\sec \theta =\frac{1}{x}\] \[\Rightarrow \tan \theta =\sqrt{{{\sec }^{2}}\theta -1}=\sqrt{\frac{1}{{{x}^{2}}}-1}=\frac{1}{\left| x \right|}\sqrt{1-{{x}^{2}}}\] Now, \[\sin {{\cot }^{-1}}\tan \theta =sinco{{t}^{-1}}\left( \frac{1}{\left| x \right|}\sqrt{1-{{x}^{2}}} \right).\] Again, putting \[x=\sin \theta ,\]we get \[\sin {{\cot }^{-1}}\left( \frac{1}{\left| x \right|}\sqrt{1-{{x}^{2}}} \right)=\sin {{\cot }^{-1}}\left( \frac{\sqrt{1-{{\sin }^{2}}\theta }}{\sin \theta } \right)\] \[=\sin {{\cot }^{-1}}\left| \cot \theta \right|=\sin \theta =x.\]
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