Show that all normals to $\gamma(t)=(\cos(t)+t\sin(t),\sin(t)-t\cos(t))$ are the same distance from the origin. Show that all normals to $\gamma(t)=(\cos(t)+t\sin(t),\sin(t)-t\cos(t))$ are the same distance from the origin. 
My attempt: 
Let $\vec{p}=(\cos(t_0)+t_0\sin(t_0),\sin(t_0)-t_0\cos(t_0))$ be any arbitrary point for $t_0\in\mathbb{R}$. Then the tangent vector at $\vec{p}$ is given by $\dot\gamma(t_0)=(t_0\cos(t_0),t_0\sin(t_0))\implies$ the slope of the tangent vector at any point is given by $m=\tan(t_0)\implies$ the slope of any normal line is given by $m_{\perp}=-\cot(t_0)$. Now we calculate the normal line at any point $\vec{p}:$ $$y-(\sin(t_0)-t_0\cos(t_0))=-\cot(t_0)(x-\cos(t_0)-t_o\sin(t_0))\implies$$ $$\cot(t_0)x+y+(2t_0\cos(t_0)+\cot(t_0)\cos(t_0)-\sin(t_0))=0$$
Recall that the distance from $Ax+By+C=0$ and $Q(x_0,y_0)$ is: $$|l,Q|=\frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}$$
Hence $$|l,Q|=\frac{\sqrt{4t_0^2\cos^2(t_0)+\cot^2(t_0)\cos^2(t_0)+\sin^2(t_0)}}{\sqrt{\cot^2(t_0)+1}}$$
How can I proceed from here? Thanks in advance!
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Further progress: 
Following the advice of user429040, the parametric form of any normal line is: $$\mathscr l=(x(t_0)-tt_0\sin(t_0), y(t_0)+tt_o\cos(t_o))$$
The goal is to now minimize the norm of this parametric line over $t$, and show that this minimum does not depend on $t_0$: 
$$|\mathscr l|=|(x(t_0)-tt_0\sin(t_0), y(t_0)+tt_o\cos(t_o))|=((t-1)^2(t_0^2+1))^\frac{1}{2}\implies\min|\mathscr l|=0.$$
However, we can graphically confirm and confirm by Prof. Blatter's answer above that this conclusion is incorrect. Where do I go from here?

As a variant: The point $\gamma(t)$ is obtained by taking the point on the circle $r(t)=(\cos(t), \sin(t))$ and following circle's tangent line at $r(t)$, which we denote by $L$, and which is in the direction $Rot_{90} r(t)=(\sin(t), -\cos(t))$, for time $t$. Thus after we compute the tangent to $\gamma(t)$ to have direction parallel to $(\cos(t), \sin(t))$ i.e. to $r(t)$, we know that the normal to $\gamma$  at $\gamma(t)$ is parallel to $L$ and passes through $\gamma(t)$, which $L$ also does. So the normal line is nothing but $L$ itself. Of course $L$ is tangent to the unit circle (at $r(t)$), and so is unit distance from the origin. 
