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If $\theta$ is in the interval $\left(0, \frac{\pi}{2}\right)$ satisfying the equation $\cos 2 \theta \cdot \sec ^4 \theta+\sec ^2 \theta=0$, then $\sin ^2 \theta=$
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$\frac{2}{3}$
$\begin{array}{lrl}\text {Given, } \cos 2 \theta \cdot \sec ^4 \theta+\sec ^2 \theta=0, \theta \in(0, \pi / 2) \\ \Rightarrow & \left(1-\tan ^2 \theta\right) \sec ^2 \theta+\sec ^2 \theta=0 \\ \Rightarrow & \sec ^2 \theta\left[1-\tan ^2 \theta+1\right]=0 \\ \Rightarrow & \tan ^2 \theta=2 \\ \Rightarrow & \sin ^2 \theta=2 / 3\end{array}$
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