<s>[INST] Simplifying $\sqrt{(\cos(\arctan x)+x\sin(\arctan x))^2-1}$ 
The expression $\sqrt{(\cos(\arctan x)+x\sin(\arctan x))^2-1}$, where $0<x<1$, is equal to $x$ or $\sqrt{(1+x^2)}$ or $\frac1{\sqrt{(1+x^2)}}$ or $\frac x{\sqrt{(1+x^2)}}$? (one of these 4 is correct).

My attempt: $$0<x<1\implies 0<\arctan x<\frac{\pi}{4}\implies\frac{1}{\sqrt2}<\cos(\arctan x)<1$$ and $$0<x\sin(\arctan x)<\frac{x}{\sqrt2}$$$$\implies\frac{1}{\sqrt2}<\cos(\arctan x)+x\sin(\arctan x)<1+\frac{x}{\sqrt2}$$$$\implies\frac{1}{2}<(\cos(\arctan x)+x\sin(\arctan x))^2<(1+\frac{x}{\sqrt2})^2$$$$\implies\sqrt\frac{-1}{2}<\sqrt{(\cos(\arctan x)+x\sin(\arctan x))^2-1}<\sqrt{(1+\frac{x}{\sqrt2})^2-1}$$$$\implies\sqrt\frac{-1}{2}<\sqrt{(\cos(\arctan x)+x\sin(\arctan x))^2-1}<\sqrt{\frac{x^2}{2}-\sqrt2x}$$ Not sure if I have reached anywhere.
 [/INST] Let $\arctan x=y;x=\tan y, -\dfrac\pi2 <y<\dfrac\pi2$ 
As $\cos y>0,$
$$\cos y=+\dfrac1{\sqrt{1+x^2}};\sin y=\tan y\cdot\cos y=\dfrac x{\sqrt{1+x^2}}$$
$$\cos(\arctan x)+x\sin(\arctan x)=\sqrt{1+x^2}$$
$\sqrt{(\cos(\arctan x)+x\sin(\arctan x))^2-1}=\sqrt{{(\sqrt{1+x^2}})^2-1}=\sqrt{x^2}=|x|$ for real $x$
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