If$A \leq_T B$ is given, can you reduce $\overline{A}$ to $B$ and vice-versa

If you are given two languages $$A$$, $$B$$ and $$A \leq_T B.$$ Is it possible to $$\overline{A} \leq_T B$$ or $$A \leq_T \overline{B}$$?

Here is my shot.

Case 1: $$\overline{A} \leq_T B$$

This is only possible if $$A \leq_m B$$ exists and $$B=\overline{B}$$. As you can transform any many-one reduction to its complement, we can show that if $$A \leq_m B$$, then $$\overline{A} \leq_m \overline{B}=B$$. Thus $$\overline{A} \leq_m B$$.

Case 2: $$A \leq_T \overline{B}$$

This is the same as above but we need to change the role of $$A$$ and $$B$$.

This is only possible if $$A \le_m B$$ exists and $$B=\overline{B}$$
But that condition simplifies to "false" since it can never the case that $$B=\overline{B}$$.
Anyway the answer to both cases is "yes". If $$A \le_T B$$ then there is a Turing machine $$T$$ with oracle $$B$$ that decides $$A$$. By complementing $$T$$'s output we obtain a Turing machine with oracle $$B$$ that decides $$\overline{A}$$ showing that $$\overline{A} \le_T B$$. Moreover, consider a Turing machine $$T'$$ with oracle $$\overline{B}$$ that simulates $$T$$ except for the following: whenever $$T$$ invokes its oracle for $$B$$ on some input $$w$$, $$T'$$ invokes its oracle for $$\overline{B}$$ on $$w$$ and then complements the result. Clearly $$T'$$ still decides $$A$$, thus showing that $$A \le_T \overline{B}$$.