= Solution
The two formulas for the <adjunction> transposes give both <triangle identities for an adjunction> directly. Apply the inverse <bijection> to the transpose of $1_{FA}$:
$$
1_{FA}=\Phi^{-1}_{A,FA}(\eta_A)=\varepsilon_{FA}F(\eta_A).
$$
Apply the forward <bijection> to the inverse transpose of $1_{GB}$:
$$
1_{GB}=\Phi_{GB,B}(\varepsilon_B)=G(\varepsilon_B)\eta_{GB}.
$$
These are equalities of components of <natural transformations>, so
$$
\boxed{(\varepsilon F)\circ(F\eta)=1_F,\qquad
(G\varepsilon)\circ(\eta G)=1_G.}
$$
No cancellation assumption on either <functor> is needed: both identities follow from the inverse <hom-set> bijections.
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