Simplification of Inverse Trigonometric and Algebraic Expression

To simplify the given expression for $0 < x < 1$:
$E = \sqrt{1+x^2}\left[\left\{x \cos \left(\cot^{-1} x\right)+\sin \left(\cot^{-1} x\right)\right\}^2-1\right]^{\frac{1}{2}}$

Let $\theta = \cot^{-1} x$. Since $0 < x < 1$, we have $\theta \in \left(\frac{\pi}{4}, \frac{\pi}{2}\right)$ and:
$\cot \theta = x$

From the trigonometric definitions for an acute angle $\theta$:
$\cos \theta = \frac{x}{\sqrt{1+x^2}}$
$\sin \theta = \frac{1}{\sqrt{1+x^2}}$

Now, evaluate the term inside the curly braces:
$x \cos \left(\cot^{-1} x\right) + \sin \left(\cot^{-1} x\right) = x \cos \theta + \sin \theta$
$= x \left(\frac{x}{\sqrt{1+x^2}}\right) + \frac{1}{\sqrt{1+x^2}} = \frac{x^2+1}{\sqrt{1+x^2}} = \sqrt{1+x^2}$

Substitute this back into the expression inside the brackets:
$\left\{x \cos \left(\cot^{-1} x\right)+\sin \left(\cot^{-1} x\right)\right\}^2 - 1 = \left(\sqrt{1+x^2}\right)^2 - 1 = (1+x^2) - 1 = x^2$

Now, taking the power of $\frac{1}{2}$:
$\left[\left\{x \cos \left(\cot^{-1} x\right)+\sin \left(\cot^{-1} x\right)\right\}^2-1\right]^{\frac{1}{2}} = \left(x^2\right)^{\frac{1}{2}} = |x|$

Since $0 < x < 1$, $|x| = x$.

Finally, multiplying by the pre-factor $\sqrt{1+x^2}$:
$E = \sqrt{1+x^2} \cdot x = x\sqrt{1+x^2}$

Therefore, the correct option is C.