= Solution
For a time-dependent <type-two generating function for a canonical transformation>, the differential identity
$$
p\,dq-H\,dt=P\,dQ-K\,dt+d(F_2-PQ)
$$
gives $K=H+\partial_tF_2$, with the <partial derivative> taken at fixed $(q,P)$. Here $F_2=\cos q\sin P$ has no explicit time dependence, so the <Hamiltonian> correction is
$$
\boxed{\Lambda(Q,P,t)=0,\qquad
K(Q,P,t)=H\left(\arccos\frac{Q}{\cos P},
-\sin P\sqrt{1-Q^2\sec^2P},t\right).}
$$
As always, adding a function of time alone to the new <Hamiltonian> leaves <Hamilton's equations> unchanged. Choosing a time-independent <type-two generating function for a canonical transformation> fixes that irrelevant freedom and gives the displayed $\Lambda=0$.
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