<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] Your function at $x=0$ is $0$. Thus it can only be $x$ or $x/\sqrt{1+x^2}$. As $x\to 1$ it tends to $1$. Hence, it must be $x$.
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