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If $\sin ( A - B ) = \frac{1}{2}$,$\cos ( A + B ) = \frac{1}{2}$,$0^{\circ} < A + B \leq 90^{\circ}$,and $A > B$,find $A$ and $B$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $m \angle A = 60^{\circ}$. If $AC = 10$,then $AB = \dots$

If $0 < \theta < 90$ and $\sin \theta = \cos 30$,then $2 \tan^2 \theta - 1 = \dots$

For $0 < \theta < 90^{\circ},$ the value of $\ldots \ldots \ldots \ldots$ increases as $\theta$ increases from $0^{\circ}$ to $90^{\circ}$.

$0 < \theta < 90$ and $\sec \theta = \operatorname{cosec} 60^\circ$,then the value of $2 \cos^2 \theta - 1$ is ........

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