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$\sin ^2 \frac{2 \pi}{3}+\cos ^2 \frac{5 \pi}{6}-\tan ^2 \frac{3 \pi}{4}=$

If $\theta$ and $\phi$ are acute angles satisfying $\sin \theta = \frac{1}{2}$ and $\cos \phi = \frac{1}{3}$,then $\theta + \phi \in$

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$\frac{1+\tanh \frac{x}{2}}{1-\tanh \frac{x}{2}}$ is equal to

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