We are given the $\ell_1$-constrained convex set $\mathcal{C} = \{ x \in \mathbb{R}^n : \| x \|_1 \leq 1 \}$, involved in a convex optimization problem. Moreover, we know that the optimal solution $x^{\star}$ has support $S := \{ i \in [n]: x^{\star}_i \neq 0 \}$.
Define the tangent cone of a convex set at a point $z$ to be the set
$$ \mathcal{K}_{\mathcal{C}}(z) = \{ \Delta \in \mathbb{R}^n: \Delta = t(x - z), \; \text{for some } t \geq 0, x \in \mathcal{C} \} $$
Question: What is the tangent cone to $\mathcal{C}$ at $x^{\star}$?
In [1], the expression for the tangent cone is given by
$$ \mathcal{K}(x^{\star}) = \{ \Delta \in \mathbb{R}^n : \langle \Delta_S, \hat{z}_S \rangle + \| \Delta_{S^c} \|_1 \leq 0 \} $$
where $\Delta_S$ is the restriction of $\Delta$ to the support set $S$ and $\hat{z}_S$ is the sign vector $\hat{z}_S := \mathrm{sign}(x^{\star}_S)$, but I can't see why. I know that the tangent cone is the polar of the normal cone at $x^{\star}$, but I can't see how this can help us reach the expression shown above.
If we try using the limiting definition of a tangent cone, we end up with the following expression:
$$ \mathcal{K}(x^{\star}) = \left\{ \Delta \in \mathbb{R}^n : x^{\star} + t \Delta \in \mathcal{C} \right\}, \forall t \geq 0 $$
That easily translates to
$$ \left\| x^{\star}_S + t \Delta_S \right\|_1 + t \left\| \Delta_{S^c} \right\|_1 \leq 1, $$
which I'm unable to further simplify. Any help?