Difference of matrix inverses into a product

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I came accross the following equality $$ (B-zI)^{-1}-(B+\tau qq^*-zI)^{-1}=\tau(B-zI)^{-1}qq^*(B+\tau qq^*-zI)^{-1} $$ Here, $z=u+iv \in \mathbb{C^+}$, $A$ is a $N\times N$ matrices, $B$ is $N\times N$ Hermitian, $\tau\in \mathbb{R}$, $q\in \mathbb{C}^N$.

Why is the above equility hold?

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Hint

Let $$A=(B-zI)^{-1}-(B+\tau qq^*-zI)^{-1}$$

Calculate $$(B-zI)A(B+\tau qq^*-zI)$$