Find `sin ""(x)/(2), cos "" (x)/(2) and tan "" (x)/(2)` in each of the case:
`tan x = (-4)/(3) , ` x in II quadrant.

Answer

To solve the problem of finding \( \sin \frac{x}{2} \), \( \cos \frac{x}{2} \), and \( \tan \frac{x}{2} \) given that \( \tan x = -\frac{4}{3} \) and \( x \) is in the second quadrant, we can follow these steps:

### Step 1: Identify the given information
We know that:
- \( \tan x = -\frac{4}{3} \)
- \( x \) is in the second quadrant.

### Step 2: Find \( \sec^2 x \)
Using the identity \( \sec^2 x = 1 + \tan^2 x \):
\[ \tan^2 x = \left(-\frac{4}{3}\right)^2 = \frac{16}{9} \]
Thus,
\[ \sec^2 x = 1 + \frac{16}{9} = \frac{25}{9} \]

### Step 3: Find \( \sec x \) and \( \cos x \)
Taking the square root gives us:
\[ \sec x = \pm \frac{5}{3} \]
Since \( x \) is in the second quadrant, \( \cos x \) is negative:
\[ \cos x = \frac{1}{\sec x} = -\frac{3}{5} \]

### Step 4: Find \( \sin x \)
Using the identity \( \sin^2 x + \cos^2 x = 1 \):
\[ \sin^2 x + \left(-\frac{3}{5}\right)^2 = 1 \]
\[ \sin^2 x + \frac{9}{25} = 1 \]
\[ \sin^2 x = 1 - \frac{9}{25} = \frac{16}{25} \]
Thus,
\[ \sin x = \sqrt{\frac{16}{25}} = \frac{4}{5} \]
Since \( x \) is in the second quadrant, \( \sin x \) is positive.

### Step 5: Find \( \cos \frac{x}{2} \)
Using the half-angle formula:
\[ \cos \frac{x}{2} = \sqrt{\frac{1 + \cos x}{2}} = \sqrt{\frac{1 - \frac{3}{5}}{2}} = \sqrt{\frac{\frac{2}{5}}{2}} = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5} \]

### Step 6: Find \( \sin \frac{x}{2} \)
Using the half-angle formula:
\[ \sin \frac{x}{2} = \sqrt{\frac{1 - \cos x}{2}} = \sqrt{\frac{1 + \frac{3}{5}}{2}} = \sqrt{\frac{\frac{8}{5}}{2}} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5} \]

### Step 7: Find \( \tan \frac{x}{2} \)
Using the relationship \( \tan \frac{x}{2} = \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}} \):
\[ \tan \frac{x}{2} = \frac{\frac{2\sqrt{5}}{5}}{\frac{\sqrt{5}}{5}} = 2 \]

### Final Results
Thus, we have:
\[ \sin \frac{x}{2} = \frac{2\sqrt{5}}{5}, \quad \cos \frac{x}{2} = \frac{\sqrt{5}}{5}, \quad \tan \frac{x}{2} = 2 \]