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

Let \(\cot ^{-1} x=\theta\), then \(x=\cot \theta\) and
\(\begin{aligned}
& \frac{\pi}{4} \lt \theta \lt \frac{\pi}{2} \quad \ldots[\because 0 \lt x \lt 1 \Rightarrow 0 \lt \cot \theta \lt 1] \\
\therefore \quad & \sqrt{1+x^2}\left[\left\{x \cos \left(\cot ^{-1} x\right)+\sin \left(\cot ^{-1} x\right)\right\}^2-1\right]^{\frac{1}{2}} \\
= & \sqrt{1+x^2}\left[\{\cot \theta \cos \theta+\sin \theta\}^2-1\right]^{\frac{1}{2}} \\
= & \sqrt{1+x^2} \cdot\left\{\left(\frac{\cos ^2 \theta+\sin ^2 \theta}{\sin \theta}\right)^2-1\right]^{\frac{1}{2}} \\
= & \sqrt{1+x^2} \sqrt{\operatorname{cosec}^2 \theta-1} \\
= & \sqrt{1+x^2}|\cot \theta| \\
= & \left(\sqrt{1+x^2}\right)|x| \\
= & x \sqrt{1+x^2}
\end{aligned}\)
\(\ldots[\because 0 \lt x \lt 1]\)