= Solution
On the interior of the principal inverse-cosine branch, $0<q<\pi$, the given <canonical transformation> has
$$
Q=\cos q\cos P,\qquad
\sqrt{1-Q^2\sec^2P}=\sin q,\qquad
p=-\sin q\sin P.
$$
Integrating $Q=\partial_PF_2$ suggests $F_2=\cos q\sin P+C(q)$. Its other derivative is $\partial_qF_2=-\sin q\sin P+C'(q)$; comparison with $p$ requires $C'(q)=0$. Hence a suitable <type-two generating function for a canonical transformation> is
$$
\boxed{F_2(q,P)=\cos q\sin P.}
$$
Both derivative equations reproduce the specified transformation exactly. Its mixed derivative is $F_{2,qP}=-\sin q\cos P$, so it defines a local <canonical transformation> where $0<q<\pi$ and $\cos P\ne0$, equivalently where $|Q/\cos P|<1$. At $q=0,\pi$ or $\cos P=0$, the stated formulas fail to give a regular coordinate chart. These restrictions explain the square-root branch and the local meaning of canonicity. This is a <trigonometric type-two canonical transformation>.
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