A) x
B) \[\frac{\pi }{2}\]
C) 1
D) None of these
Correct Answer: A
Solution :
Let \[{{\cos }^{-1}}x=\theta \,\,\,\Rightarrow \,\,x=\cos \theta \,\,\Rightarrow \,\,\sec \theta =\frac{1}{x}\] \[\Rightarrow \,\,\tan \theta =\sqrt{{{\sec }^{2}}\theta -1}=\sqrt{\frac{1}{{{x}^{2}}}-1}=\frac{1}{x}\sqrt{1-{{x}^{2}}}\] Now \[\sin \,\,{{\cot }^{-1}}\tan \theta =\sin \,\,{{\cot }^{-1}}\,\left( \frac{1}{x}\sqrt{1-{{x}^{2}}} \right)\] Again, putting \[x=\sin \theta \] \[\sin \,\,{{\cot }^{-1}}\left( \frac{1}{x}\sqrt{1-{{x}^{2}}} \right)=\sin \,\,{{\cot }^{-1}}\left( \frac{\sqrt{1-{{\sin }^{2}}\theta }}{\sin \theta } \right)\] \[=\sin \,\,{{\cot }^{-1}}\,(\cot \theta )=\sin \theta =x\].
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