$\DeclareMathOperator\Sha{Sha}\DeclareMathOperator\Gal{Gal}$Let $L/K$ be a quadratic extension of number field $K$. Let $\sigma$ be a generator of $\Gal(L/K)$.

Let $E/K$ be an elliptic curve defined over $K$ and $\Sha(E/K)$ be its Tate-Shafarevich group. $\Gal(L/K)$ acts on $\Sha(E/L)$ naturally (cf. How Galois group acts on Tate-Shafarevich group?).

There is a canonical isomorphism $\tau: E(L)\cong E_D(L), (x,y)\mapsto (x,y/\sqrt{D})$ .

My goal is to prove $\tau$ induces $(1-\sigma)\text{Sha}(E/L)\cong 　\text{trace}\text{Sha}(E_D/L)$. This isomorphism appears in p219 of link)

To prove this isomorphism, it is enough to prove $\sigma \phi(C)=-\phi \sigma(C)$ for all $C \in \Sha(E/L)$.

there exists an isomorphism $\phi: \Sha(E/L)\cong \Sha(E_D/L)$ induced by $\tau$ though I cannot write down the map between them.

we can check $\sigma \tau=-\tau \sigma$ because we can calculate its coordinates explicitly, but I cannot calculate both $\sigma \phi$ and $\phi \sigma$, so I'm having difficulty to prove $\sigma \phi(C)=-\phi \sigma(C)$ for all $C \in \Sha(E/L)$. How can I overcome this trouble and prove $\sigma \phi(C)=-\phi \sigma(C)$ for all $C \in \Sha(E/L)$ ?

Another approach of proving $(1-\sigma)\text{Sha}(E/L)\cong \text{trace}\text{Sha}(E_D/L)$ is also appreciated. Thank you in advance.