Find $\sin \frac{x}{2}, \cos \frac{x}{2}$ and $\tan \frac{x}{2},$ if $\tan x = -\frac{4}{3}$ and $x$ is in quadrant $II$.

(D) Given that $x$ is in quadrant $II$,so $\frac{\pi}{2} < x < \pi$.
Dividing by $2$,we get $\frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}$.
This implies that $\frac{x}{2}$ lies in the first quadrant,so $\sin \frac{x}{2}, \cos \frac{x}{2},$ and $\tan \frac{x}{2}$ are all positive.
Given $\tan x = -\frac{4}{3}$,we find $\sec^2 x = 1 + \tan^2 x = 1 + \frac{16}{9} = \frac{25}{9}$.
Thus,$\cos^2 x = \frac{9}{25}$,which gives $\cos x = -\frac{3}{5}$ (since $x$ is in quadrant $II$,$\cos x < 0$).
Using the formula $\cos x = 2\cos^2 \frac{x}{2} - 1$,we have $-\frac{3}{5} = 2\cos^2 \frac{x}{2} - 1$.
$2\cos^2 \frac{x}{2} = 1 - \frac{3}{5} = \frac{2}{5} \Rightarrow \cos^2 \frac{x}{2} = \frac{1}{5}$.
Since $\frac{x}{2}$ is in the first quadrant,$\cos \frac{x}{2} = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5}$.
Using $\sin^2 \frac{x}{2} = 1 - \cos^2 \frac{x}{2} = 1 - \frac{1}{5} = \frac{4}{5}$.
Since $\frac{x}{2}$ is in the first quadrant,$\sin \frac{x}{2} = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5}$.
Finally,$\tan \frac{x}{2} = \frac{\sin(x/2)}{\cos(x/2)} = \frac{2/\sqrt{5}}{1/\sqrt{5}} = 2$.
Therefore,$\sin \frac{x}{2} = \frac{2\sqrt{5}}{5}, \cos \frac{x}{2} = \frac{\sqrt{5}}{5},$ and $\tan \frac{x}{2} = 2$.