If \(0 < x < 1\), 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

We have, \(0 < x < 1\)
Let \(\quad \cot ^{-1} x=\theta\)
\(\Rightarrow \quad \cot \theta=x\)
\(\therefore \quad \sin \theta=\frac{1}{\sqrt{1+x^2}}=\sin \left(\cot ^{-1} x\right)\)
and \(\quad \cos \theta=\frac{x}{\sqrt{1+x^2}}=\cos \left(\cot ^{-1} x\right)\)
Now, \(\sqrt{1+x^2}\left[\left\{x \cos \left(\cot ^{-1} x\right)+\sin \left(\cot ^{-1} x\right)\right\}^2-1\right]^{1 / 2}\)

\[
\text { Now, } \begin{aligned}
& \sqrt{1+x^2}\left[\left\{x \cos \left(\cot ^{-1} x\right)+\sin \left(\cot ^{-1} x\right)\right\}^2-1\right]^{1 / 2} \\
&=\sqrt{1+x^2}\left[\left\{x \frac{x}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+x^2}}\right\}^2-1\right]^{1 / 2} \\
&=\sqrt{1+x^2}\left[\left(\frac{1+x^2}{\sqrt{1+x^2}}\right)^2-1\right]^{1 / 2} \\
&=\sqrt{1+x^2}\left[1+x^2-1\right]^{1 / 2} \\
&=x \sqrt{1+x^2}
\end{aligned}
\]