6. As shown in Figure 2, square $A B C D$ is inscribed in $\odot O, P$ is a point on the minor arc $\overparen{C D}$, $P A$ intersects $B D$ at point $M$, $P B$ intersects $A C$ at point $N$, and let $\angle P A C=\theta$. If $M N \perp P A$, then $2 \cos ^{2} \theta-\tan \theta=(\quad)$.
(A) 1
(B) $\frac{\sqrt{2}}{2}$
(C) $\frac{1}{2}$
(D) $\frac{\sqrt{2}}{4}$
6. A.

Given that quadrilateral $A B C D$ is a square, we know
$$
\begin{array}{l}
\angle A C B=45^{\circ}, D B \perp A C \\
\Rightarrow \angle A P B=\angle A C B=45^{\circ} . \\
\text { Also, } M N \perp P A \\
\Rightarrow \angle M N P=\angle A P B=45^{\circ} \\
\Rightarrow M P=M N .
\end{array}
$$

Since $A C$ is the diameter of the circle, $\angle A P C=90^{\circ}$. Therefore, points $P, M, O, C$ are concyclic. Thus, $A M \cdot A P=A O \cdot A C$.
Hence, $2 \cos ^{2} \theta-\tan \theta=2 \cdot \frac{A O^{2}}{A M^{2}}-\frac{M N}{A M}$
$$
\begin{array}{l}
=\frac{A O \cdot A C}{A M^{2}}-\frac{M N}{A M}=\frac{A P \cdot A M}{A M}-\frac{M N}{A M} \\
=\frac{A P-P M}{A M}=1
\end{array}
$$