While I was reading the book, Discriminants, Resultants, and Multidimensional Determinants by Gelfand, Kapranov & Zelevinsky, I came upon the following statement:
Let $X \subseteq \mathbb{P}^n$ be a smooth projective variety such that $X$ is not contained in any hyperplane. Let $H \subseteq \mathbb{P}^n$ an hyperplane and $X^\vee$ the dual variety to $X$. Then the following statements are equivalent:
- $H \in X^\vee$.
- $X \cap H$ is singular (as a scheme).
My attempt on $2 \Rightarrow 1$ was to go by contradiction but I could not finish the proof. On the other hand, for $1 \Rightarrow 2$ I could only prove it in the case of plane curves by the means of the Gauss map. I would like to know how to prove both implications.