Prove $\sinh(u+v-\eta)\sinh(2v)-\sinh(\eta)\sinh(u-v) = \sinh(2v-\eta)\sinh(u+v)$

Question

How to prove ?

$$\frac{\sinh(u+v-\eta)\sinh(2v)-\sinh(\eta)\sinh(u-v)}{\sinh(u+v)} = \sinh(2v-\eta)$$

Answer 1

HINT: use that $$ \sinh z=\frac{e^{z}-e^{-z}}{2} $$