Let $R$ be a Commutative Noetherian ring and let $\textbf{R} \text{Hom}_R(-,-)$ and $-\otimes_R^{\textbf{L}}-$ denote derived Hom and tensor product respectively.
We recall that for any $R$-modules $M,N$ we have that $$\text H_{-m}(\textbf{R} \text{Hom}_R(M,N))=\text{Ext}^m_R(M,N),\forall m\in \mathbb Z$$ and $\text H_m (M\otimes_R^{\textbf{L}}N)=\text{Tor}_m^R(M,N),\forall m\in \mathbb Z$. So in particular , we have
$\text H_{m}(\textbf{R} \text{Hom}_R(M,N))=0,\forall m>0$ and $\text H_m (M\otimes_R^{\textbf{L}}N)=0,\forall m<0$.
Hence, am I correct in saying the following : If for $R$-modules $W,X,Y,Z$, we have $\textbf{R} \text{Hom}_R(W,X)\cong Y\otimes_R^{\textbf{L}}Z$, then (taking suitable homologies on both sides), we get
$\text{Ext}^m_R(W,X)=\text{Tor}_m^R(Y,Z)=0,\forall m>0.$ ?