A) \[\sqrt{1+{{x}^{2}}}\]
B) \[x\]
C) \[{{(1+{{x}^{2}})}^{-3/2}}\]
D) \[{{(1+{{x}^{2}})}^{-1/2}}\]
Correct Answer: D
Solution :
Let \[{{\cot }^{-1}}x=\theta \,\,\Rightarrow \,\,x=\cot \theta \] Now \[\cos ec\,\theta =\sqrt{1+{{\cot }^{2}}\theta }=\sqrt{1+{{x}^{2}}}\] \[\therefore \,\,\,\sin \theta =\frac{1}{\cos ec\,\theta }=\frac{1}{\sqrt{1+{{x}^{2}}}}\,\,\Rightarrow \,\theta ={{\sin }^{-1}}\frac{1}{\sqrt{1+{{x}^{2}}}}\] Hence \[\sin \,({{\cot }^{-1}}x)\,=\sin \,\left( {{\sin }^{-1}}\frac{1}{\sqrt{1+{{x}^{2}}}} \right)\] \[=\frac{1}{\sqrt{1+{{x}^{2}}}}={{(1+{{x}^{2}})}^{-1/2}}\].
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