[Solution] Let $h$ be the length of the paper. Then, in the right triangle $\triangle ABC$,
$$
\begin{aligned}
h & =\frac{6}{\cos \left(90^{\circ}-2 \theta\right)}=\frac{6}{\sin 2 \theta} \\
& =3 \sec \theta \csc \theta, \\
L & =h \sec \theta \\
& =3 \sec ^{2} \theta \csc \theta .
\end{aligned}
$$

Therefore, the answer is $(A)$.
$7 \cdot 37$ A rectangular piece of paper 6 inches wide is folded as shown in the figure so that one corner touches the opposite side. The length in inches of the crease $L$ in terms of the angle $\theta$ is
(A) $3 \sec ^{2} \theta \csc \theta$.
(B) $6 \sin \theta \sec \theta$.
(C) $3 \sec \theta \csc \theta$.
(D) $6 \sec \theta \csc ^{2} \theta$.
(E) None of these.
(23rd Annual American High School Mathematics Examination, 1972)