If \(0 \lt x \lt 1\); then
\(\sqrt{1+x^2}\left[\left\{x \cos \left(\cot ^{-1} x\right)+\sin \left(\cot ^{-1} x\right)\right\}^2-1\right]^{\frac{1}{2}}=\)

Given,
\(\sqrt{1+x^2}\left[\left\{x \cos \left(\cot ^{-1} x\right)+\sin \left(\cot ^{-1} x\right)\right\}^2-1\right]^{\frac{1}{2}}\)
Let \(\cot ^{-1} x=\theta\)
\(\begin{aligned}
\therefore \quad & x=\cot \theta \\
& \Rightarrow \sin \theta=\frac{1}{\sqrt{1+x^2}}...(i) \\
& \Rightarrow \cos \theta=\frac{x}{\sqrt{1+x^2}}
...(ii)\end{aligned}\)
\(\therefore \quad \sqrt{1+x^2}\left[\left\{x \cos \left(\cot ^{-1} x\right)+\sin \left(\cot ^{-1} x\right)\right\}^2-1\right]^{\frac{1}{2}} \)
\( =\sqrt{1+x^2}\left[\{x \cos \theta+\sin \theta\}^2-1\right]^{\frac{1}{2}} \)
\( =\sqrt{1+x^2}\left[\left\{x \times \frac{x}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+x^2}}\right\}^2-1\right]^{\frac{1}{2}} \)
\( =\sqrt{1+x^2}\left[\left(\frac{x^2+1}{\sqrt{1+x^2}}\right)^2-1\right]^{\frac{1}{2}} \)
\( = \sqrt{1+x^2}\left[\left(\sqrt{1+x^2}\right)^2-1\right]^{\frac{1}{2}} \)
\( = \sqrt{1+x^2}\left(1+x^2-1\right)^{\frac{1}{2}} \)
\( = x \sqrt{1+x^2}\)