The value of $\frac{\sec ^6 \theta-\tan ^6 \theta-3 \sec ^2 \theta \tan ^2 \theta+1}{\cos ^4 \theta-\sin ^4 \theta+2 \sin ^2 \theta+2}$ is :

1 $\frac{1}{2}$ $\frac{2}{3}$ $\frac{3}{4}$

$\frac{2}{3}$