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

If $\cot x = -\frac{5}{12}$ and $x$ lies in the second quadrant,find the values of the other five trigonometric functions.

$\operatorname{Tanh}^{-1}(\sin \theta) =$

If $\theta$ lies in the first quadrant and $5 \tan \theta = 4$,then $\frac{5 \sin \theta - 3 \cos \theta}{\sin \theta + 2 \cos \theta}$ is equal to

$\tan 1^\circ \tan 2^\circ \tan 3^\circ \tan 4^\circ \dots \tan 89^\circ = $

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