If $0<x<1$, then $\sqrt{1+x^{2}}\left[\left\{x\cos(\cot^{-1}x)+\sin(\cot^{-1}x)\right\}^{2}-1\right]^{1/2}=$

If 0<x<1, then √1+x²[{xcos(cot⁻¹x)+sin(cot⁻¹x)}²-1]^(1/2)= $0 < x < 1$ $\sqrt{1+x^{2}}\left[\left\{x\cos(\cot^{-1}x) + \sin(\cot^{-1}x)\right\}^{2} - 1\right]^{1/2}$ $\text{To find: the simplified value of the expression}$ $1.\ \text{put } \theta = \cot^{-1}x$ $2.\ \text{find } \sin\theta,\ \cos\theta \text{ from a right triangle}$ $3.\ \text{simplify the bracket, then take the root}$ $\theta = \cot^{-1}x,\quad \cot\theta = x,\quad \theta \in \left(\dfrac{\pi}{4}, \dfrac{\pi}{2}\right)$ $\sin\theta = \dfrac{1}{\sqrt{1+x^{2}}}, \qquad \cos\theta = \dfrac{x}{\sqrt{1+x^{2}}}$ $x\cos\theta = x \cdot \dfrac{x}{\sqrt{1+x^{2}}} = \dfrac{x^{2}}{\sqrt{1+x^{2}}}$ $x\cos\theta + \sin\theta = \dfrac{x^{2}}{\sqrt{1+x^{2}}} + \dfrac{1}{\sqrt{1+x^{2}}} = \dfrac{1+x^{2}}{\sqrt{1+x^{2}}}$ $x\cos\theta + \sin\theta = \sqrt{1+x^{2}}$ $\left\{x\cos\theta + \sin\theta\right\}^{2} - 1 = (1+x^{2}) - 1 = x^{2}$ $\left[x^{2}\right]^{1/2} = |x| = x \quad (\text{since } 0 < x < 1)$ $\sqrt{1+x^{2}} \cdot x = x\sqrt{1+x^{2}}$ $x = 0.75:\ \sqrt{1+x^{2}}\cdot x = 1.25 \times 0.75 = 0.9375$ $\boxed{\text{Answer: (3) } x\sqrt{1+x^{2}}}$