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$\cos ^2\left(\frac{7 \pi}{8}\right)+\cos ^2\left(\frac{5 \pi}{8}\right)+\cos ^2\left(\frac{3 \pi}{8}\right)+\cos ^2\left(\frac{\pi}{8}\right)=$

Prove that: $\cot ^{2} \frac{\pi}{6} + \csc \frac{5 \pi}{6} + 3 \tan ^{2} \frac{\pi}{6} = 6$

If $\tan \theta = \frac{\sin \alpha - \cos \alpha}{\sin \alpha + \cos \alpha}$,then $\sin \alpha + \cos \alpha$ and $\sin \alpha - \cos \alpha$ are equal to:

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

Convert $40^{\circ} 20^{\prime}$ into radian measure.

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