$\frac{x}{\sqrt{1+x^{2}}}$ $\frac{1}{\sqrt{1+x^{2}}}$ $\frac{\sqrt{1+x^{2}}}{x}$ $\sqrt{1+x^{2}}$

Let $\cot ^{-1}x = \Theta$

$\Rightarrow \cot\Theta = x$

Using Pythagoras theorem, $\sin \Theta =\frac{1}{\sqrt{1+x^2}}$

$\Rightarrow \Theta =\sin ^{-1} \left ( \frac{1}{\sqrt{1+x^2}} \right )$

$\sin \left ( \sin ^{-1} \left ( \frac{1}{\sqrt{1+x^2}} \right ) \right ) = \frac{1}{\sqrt{1+x^2}}$

Option (2) is correct

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