`int(dx)/(sin^2xcos^2x)` equals
(A) `tanx+cotx+C`
(B) `tanx-cotx+C`
(C) `tanxcotx+C`
(D) `tanx-cot2x+C`

Explanation

To solve the integral \( \int \frac{dx}{\sin^2 x \cos^2 x} \), we can start by rewriting the integrand using trigonometric identities.

### Step 1: Rewrite the integrand
We know that:
\[ \sin^2 x \cos^2 x = \frac{1}{4} \sin^2(2x) \]
Thus, we can rewrite the integral as:
\[ \int \frac{dx}{\sin^2 x \cos^2 x} = \int \frac{4 \, dx}{\sin^2(2x)} \]

### Step 2: Use the cosecant identity
Using the identity \( \csc^2(2x) = \frac{1}{\sin^2(2x)} \), we can rewrite the integral:
\[ \int \frac{4 \, dx}{\sin^2(2x)} = 4 \int \csc^2(2x) \, dx \]

### Step 3: Integrate using the known integral
The integral of \( \csc^2(kx) \) is known to be:
\[ \int \csc^2(kx) \, dx = -\frac{1}{k} \cot(kx) + C \]
In our case, \( k = 2 \):
\[ 4 \int \csc^2(2x) \, dx = 4 \left(-\frac{1}{2} \cot(2x) + C\right) = -2 \cot(2x) + C \]

### Final Result
Thus, the final result of the integral is:
\[ \int \frac{dx}{\sin^2 x \cos^2 x} = -2 \cot(2x) + C \]

### Conclusion
The correct option is not listed among the choices provided. However, if we consider the relationship between \( \tan \) and \( \cot \), we can express \( -2 \cot(2x) \) in terms of \( \tan \) and \( \cot \) functions, but it does not match any of the options directly.