If `tanx=-(4)/(3), (3pi)/(2) lt x lt 2pi`, find the value of `9sec^(2)x-4 cot x`.

Answer

To solve the problem, we need to find the value of \( 9\sec^2 x - 4\cot x \) given that \( \tan x = -\frac{4}{3} \) and \( \frac{3\pi}{2} < x < 2\pi \).

### Step 1: Identify the quadrant
Since \( \tan x = -\frac{4}{3} \) and \( x \) is in the fourth quadrant (between \( \frac{3\pi}{2} \) and \( 2\pi \)), we know that \( \tan x \) is negative in this quadrant.

### Step 2: Find \( \sec^2 x \)
Using the identity:
\[ \sec^2 x = 1 + \tan^2 x \]
First, we calculate \( \tan^2 x \):
\[ \tan^2 x = \left(-\frac{4}{3}\right)^2 = \frac{16}{9} \]
Now substituting this value into the identity:
\[ \sec^2 x = 1 + \frac{16}{9} = \frac{9}{9} + \frac{16}{9} = \frac{25}{9} \]

### Step 3: Find \( \cot x \)
We know that:
\[ \cot x = \frac{1}{\tan x} \]
Thus:
\[ \cot x = \frac{1}{-\frac{4}{3}} = -\frac{3}{4} \]

### Step 4: Substitute values into the expression
Now we substitute \( \sec^2 x \) and \( \cot x \) into the expression \( 9\sec^2 x - 4\cot x \):
\[ 9\sec^2 x = 9 \times \frac{25}{9} = 25 \]
\[ -4\cot x = -4 \times \left(-\frac{3}{4}\right) = 3 \]
Now we combine these results:
\[ 9\sec^2 x - 4\cot x = 25 + 3 = 28 \]

### Final Answer
Thus, the value of \( 9\sec^2 x - 4\cot x \) is \( \boxed{28} \).