(D) Given $\sin A = \frac{4}{5}$ and $90^\circ < A < 180^\circ$ (second quadrant).Since $\sin A$ is positive and $\cos A$ is negative in the second qu

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(D) Given $\sin A = \frac{4}{5}$ and $90^\circ Since $\sin A$ is positive and $\cos A$ is negative in the second quadrant,$\cos A = -\sqrt{1 - \sin^2 A} = -\sqrt{1 - (\frac{4}{5})^2} = -\frac{3}{5}$.We know that $\tan \frac{A}{2} = \sqrt{\frac{1 - \cos A}{1 + \cos A}}$.Since $90^\circ $\tan \frac{A}{2} = \sqrt{\frac{1 - (-3/5)}{1 + (-3/5)}} = \sqrt{\frac{1 + 3/5}{1 - 3/5}} = \sqrt{\frac{8/5}{2/5}} = \sqrt{4} = 2$.